Even the earthquake is spreading in waves, so the peak acceleration hits the lights sequentially, one after one. So even with that the lamps never break many at once, the time differences could be in 10's of ms for the shunts to activate.

What I've seen on a few pictures, the cutouts were made of some kind of hardened paper, so very brief overvoltage is sufficient to ignite the arc through the paper and carbonize that area within few ms. And when carbonized, it becomes at least somehow conductive, so it allows breakdown of the discs as well. Then the resistive nature of the carbonized paper makes it heat so it burns out, so allows the contacts completely to short circuit, but that does not need the elevated voltage anymore.

That is, why i noted them breaking in day time when there is no power. Then when the night comes the power is applied at once

Since there are leakage resistance to Earth from each of the "floating" sections of the circuit, i would expect near full OCV to appeaer across the isolator at the high end, so there is one from which the reaction starts. But, would such discharge (into leakage, so very small current) even be sufficient to destroy the isolator ?

In case with the paper, some initial carbonization will make it drop in resistance from "infinity" to some high value. As it is getting to the same order of magnitude as the leakage to Earth, the voltage across it goes down, and across the next one goes up, so same process repeat with the next one while it slows down with the 1st. I wonder if it may happen that they all go into some high resistance value that is about balanced, and get "stuck" in this condition for some time (minutes, hours) before carbonization advances sufficiently to make for a final breakdown in one, after which the rest will follow

I assume there is also a fuse that will blow if too many burnt out lamps are left installed and with shunts activated. I don't imagine the shunts to offer the same resistance as the lamps so the current in the other still working lamps would be higher and higher as more lamps burn out and the shunts activate. When too many shunts have activated, the current goes over the fuse limit and it blows.

When too many shunts have activated, the current goes over the fuse limit and it blows.

These use to be fed from a constant current transformer , which deliver the same current from zero volts till the full voltage of all lamps. So all what happens while more bulbs get broken is, the secondary winding in the transformer is moving further away from the primary...

I think there would be very little leakage to earth after all any leakage like that presents a shock hazard and the power company's don't want that on there minds. I think the paper shunts are treated chemically so that when a voltage is applied when the filament fails it quickly burns up then 2 wire touch and complete an almost zero ohms link to keep things running

The leakage does not pose any hazard to who is standing on the Earth

Leakages are a fact in any electrical grid. They are not a fault, it is simply not possible to build a grid without leakage. At voltages in the 100s-V or few-kV range i'd expect the leakage current will be sufficient to destroy breaking-down isolation in series with it

Paper ot its own is fair enough for this task. It burns through and the metal contacts touch directly....Though i would expect there something like a tab made of clay or such, that would shatter from the heat of the arc when it breaks down

If I understand correctly, the transformer adjusts itself to deliver the correct current to all lamps by lowering the voltage of the entire loop. This means all lamps see the voltage and current they were designed for regardless of the number of EOL lamps. There is no need for a fuse that has the same purpose as the one in Christmas lights, and other lamps can still remain in the circuit when a number of them have blown. It even compensates enough to light a single lamp if all others are bad.

In large yep thats how it is

I think there may be some minimum load requirement though (ie. minimum voltage that must drop on the loop). To get working current regulation, the secondary armature must be floating above the primary, so f = mg. The higher load, higher magnetic field produced, so the point where f = mg is further up away from the primary (where mg is independent of the load). But with too low load, f may be too low to lift the armature up even when it is in the bottom position....

There is indeed maximum voltage limit, where exceeding that means the current would drop.
But with near short circuit there is always the same current, so the same force lifting the winding (or it's counterweight - depend in which position it is supposed to operate).
The transformer regulates the current in fact by varying the leakage inductance between primary and secondary. So higher the voltage difference between the OCV and load voltage, less movement is necessary when the load voltage changes, when the load is the range of 25% or lower, the winding won't move at all and yet the current stays within 5% even till a short circuit...
Of course, a difference is, when the secondary power (so voltage) is boosted by an extra fixed transformer (extra voltage in series with the secondary), this extra voltage might be, what pushes the moving coil till the end of it's regulation range at short circuit, then the current may become higher than rated.