I got a couple 18650 cells from a laptop pack, they seem to be still alive but need to be charged, but I don't have a proper charger. What other methods can I use? Can I change them with 5V DC with a resistor in series? Or a capacitive dropper? I don't have a power bank that operates in those cells to change them. The batteries show around 2V average.

Do not charge without proper charging AND protection circuits.
You would need to limit the voltage to 4.2V +/- 1%, exceeding about 4.3V means you are asking for quite nasty fireworks with smoke effects and maybe an explosion. So you need a redundant system, where the 4.2V charger control has a backup protection circuitry designed to disconnect the battery even when the charger fails.

You may buy charger/protection modules on practically all the popular sites for ferw cents a piece, look for "tp4056 usb 1a liion charger board".
Two types use to be there, one with protection and one just with the charger chip. Use the one with protection.
I would buy a pack of 5 or so and connect each together with its cell and tie it together (maybe with a switch in the output leads). That way you may get selfcontained 3.6V source for your experiments, with micro-USB charging connector and protected against short circuits on the output...

Now I replaced the batteries from an old camcorder pack with the ones from the laptop. They run in series and the pack have a circuit, both batteries in series show around 4V before the circuit and 0V at the output of the circuit, probably an over discharge protection. Should I stick directly 7.5V to the battery pack though the circuit?

Edit: I have an old camcorder that I want to power up somehow to see if it works, otherwise to take it into bits and take some nice things like a mini CRT TV, motors and many juicy things.

Now I replaced the batteries from an old camcorder pack with the ones from the laptop. They run in series and the pack have a circuit, both batteries in series show around 4V before the circuit and 0V at the output of the circuit, probably an over discharge protection. Should I stick directly 7.5V to the battery pack though the circuit?

Edit: I have an old camcorder that I want to power up somehow to see if it works, otherwise to take it into bits and take some nice things like a mini CRT TV, motors and many juicy things.

The circuit in the battefy pack is just the protection part (not the charger control itself), but most likely (assume it is tge same as the notebook battery packs) it is not designed to have the cells replaced. Point is, it likely has, as part of the protection, a special fuse with "trip" connection, which the battery supervisor triggers once it sees the batteries at EOL (so cells prone to fire/explosion even with a normal charger), disconnecting it for good, so ensuring the bad batterieas would never get charged again. It is very likely the owner has just used the dying battery pack till really the point the supervisor decided to fire the fuse. In that case it is practically over and the protection board is just dead.
Many designs program an internal EEPROM code to never ever allow to charge the cells again. Plus to avoid nuisance tripping, the condition triggering the fuse trip use to be quite complex, involving "observation" of tye history of the battery pack. And there even when you replace the fuse along with the cells, the controller would still think it has the old, worn out cells (and would misdiagnose the suddenly increased charge the cells need to be fully charged as a severe internal leakage so extreme fire risk) and would likely trip the new fuse again. Plus it will keep the disconnect mosfets off too...
As these controllers are pretty custom software, there is no realistic way to reset them to the "fresh cell" state. Plus because the EEPROM has a limited life too (it gets constantly rewritten as the cell life state data is regularly updated there), with new cells it could be the EEPROM tyat will fail first, leading to the fuse firing (good designs will have some data consistency detection and bad data = potencially dangerous => the only safe thing is to fire the fuse) or with shoddy design the protection module failing to prevent the battery runaway.

I've doing experiments with it and I figured how to change the batteries "safely". I have a boombox that I took apart that works with two 18650s in series, it is has a 9V input with a resistive dropper and an charging indication LED. When the battery is charging (drawing current from the circuit) it turns red, when the battery is charged (when it reaches the cutout voltage, no current drawing) it turns blue.

The curious thing is that when I try to charge the batteries through the original protection circuit with the boombox the light turns red, and it draws 0.3A, but the battery nor the protection circuit are hot, only the resistive dropper and the switching power supply, so the charging stage is ok and the circuit doesn't decide to kill the charge yet, that's weird because in leaked or damaged batteries this could result in nasty overheating and hydrogen production. But when I disconnect it from the charger I see no voltage at the output of the circuit, I stick a lamp load to simulate the camera but nothing, it could be that I haven't charged the batteries enough or the circuit decided to kill the output only allowing charge.

Could the protection chip be detecting when the battery pack is being hacked by changing the batteries? I think I can't replace the original control board for a generic one (which is likely to not have programmed obsolescence, as it is supposed go around between many batteries), but the camera have a control or communication port to the battery.

Edit 2: The reverse polarity protection seems to be working, I accidentally connected the polarity backwards to the charger and the light keeps blue.

I've doing experiments with it and I figured how to change the batteries "safely". I have a boombox that I took apart that works with two 18650s in series, it is has a 9V input with a resistive dropper and an charging indication LED. When the battery is charging (drawing current from the circuit) it turns red, when the battery is charged (when it reaches the cutout voltage, no current drawing) it turns blue.

It should have a charging controller as well (a small sot23 transistor size IC, the extra dropping resistor is then used to prevent power being dissipated on that IC). But double check, what is the cutout voltage (should be 8.4V; if you see something like 8.7V or more, it is just the protection and it is not safe).

The curious thing is that when I try to charge the batteries through the original protection circuit with the boombox the light turns red, and it draws 0.3A, but the battery nor the protection circuit are hot, only the resistive dropper and the switching power supply, so the charging stage is ok and the circuit doesn't decide to kill the charge yet, that's weird because in leaked or damaged batteries this could result in nasty overheating and hydrogen production. But when I disconnect it from the charger I see no voltage at the output of the circuit, I stick a lamp load to simulate the camera but nothing, it could be that I haven't charged the batteries enough or the circuit decided to kill the output only allowing charge.

Nor the batteries, nor the protection should become any hot at all. If they are, something is really bad there.
It could have been the overdischarge protection has triggered. In that case it may need the full charge cycle to reset (and remeasure the battery capacity)

Could the protection chip be detecting when the battery pack is being hacked by changing the batteries? I think I can't replace the original control board for a generic one (which is likely to not have programmed obsolescence, as it is supposed go around between many batteries), but the camera have a control or communication port to the battery.

Edit 2: The reverse polarity protection seems to be working, I accidentally connected the polarity backwards to the charger and the light keeps blue.

It does not explcitly detect the cells being replaced, but it may get confused by the sudden parameter change and so assume it is a fault (an excessive leakage which leads to fireworks exhibits the same way as when the battery capacity has increased due to cells being replaced: In both cases the voltage is rising slower than it did the previous cycles the controller remembers in its EEPROM; as the design assumes no one will be replacing the cells, it considers that signature as a sign of a dangerous battery fault).
Using other controller board may mean two things:
The other board may be explicitly designed for the batteries to be replaced, so may feature extra mechanisms able to safely distinguish between replaced vs failing cells (or need some programming to be done to reset it). But with cheepeese boards from AliExpress there is extremely high risk these boards would just skip such fault detecting features and leave the cells a ticking fireworks bomb.
Definitely it is not just the batteries, what is wearing out, it is the protection IC as well. And it is then a question, if the worn out IC wont shut down the pack soon after the cells were replaced because of data corruption in its EEPROM.

Some controllers (those using just RAM and relying on at least some cells retaining the voltage sufficient for data retention) do not wear out and are reset by replacing the cells, but then you absolutely must connect the new ones in the exact specified order. And the cells must be replaced before the controller detect their degradation as unsafe to charge anymore, because then it always triggers the fuse.
Connecting the new cells in a wrong order may cause the controller to misidentify the voltage condition as a battery failure and fire the fuse. Often the correct order is to start with the most negative cell and only then connect the rest.
But these controllers have to fire the fuse way sooner than those with EEPROM, so usually are not that popular anymore. But it could be the case in your camera.

Now I charged the battery completely (until the cutout voltage of the "charger", which was around 8.2V), and now the circuit showed 8.2V at the output, so it was an over discharge protection, I've also accidentally connected the polarity backwards and no crazy current draw from the charger or fireworks, so the reverse polarity protection is working. I've also connected a 21W 12V lamp at the output of the circuit and it didn't light up and it remains 0V until you charge it again. So seems like the circuit got reset when I changed the batteries (the program may be written in ROM but the cycle counts in RAM), or there is still plenty of cycles more until the circuit decides to kill the battery.

I was trying to make it work for power up an old camcorder, but the camcorder was completely dead and so I took it into bits, and I won't do more with that battery pack, I could probably use it to charge more batteries or use it in another camcorder.

Now I charged the battery completely (until the cutout voltage of the "charger", which was around 8.2V), and now the circuit showed 8.2V at the output, so it was an over discharge protection, I've also accidentally connected the polarity backwards and no crazy current draw from the charger or fireworks, so the reverse polarity protection is working. I've also connected a 21W 12V lamp at the output of the circuit and it didn't light up and it remains 0V until you charge it again. So seems like the circuit got reset when I changed the batteries (the program may be written in ROM but the cycle counts in RAM), or there is still plenty of cycles more until the circuit decides to kill the battery.

The protection does not count cycles, as with LiIon the number of cycles are nearly irrelevant to the wear. What is way more relevant is the correlation between cell voltage and temperature extremes and that is nearly impossible to track, plus the response is quite unpredictable
It usually monitors the voltage and tracks the real capacity the cells still have in them, their internal resistance, the degree of the required balancing actions (how much balancing correction the pack needed during charging) and leakage mainly during charging. Once these parameters exceed preprogrammed EOL limits, the controller declares the pack as EOL and prevents further charging (some fire the fuse at this point t really disconnect the battery). It is programmed with an assumption the cells will normally degrade in these parameters just gradually and most only in one way (the lost capacity won't recover by itself, but the cell delamination may appear as the capacity changing back and forth).

In your case it seems the cells hadn't reached the state to cause the protection circuit to trip, which is good for you.
The more thorough protections (those not relying on RAM, but using EEPROM instead, so impossible to reset) came after the wave of battery fire incidents all around the world (about a decade ago, I think Apple was quite affected there), so some older equipment may just not be using it yet.