I do not know what the properties of this power supply are - whether it has output voltage stabilization or not. If not, then plugging in a higher voltage battery likely created higher voltage and breakdown - however, usually the breakdown reserve is assumed to be large enough that if the silicone were new it would not have happened. But if it was old, if there were cracks in it, it could have happened - especially cracks, even microscopic ones, are of great importance because they make the electric field locally stronger (just like a blade).
There are several readily available materials that are good insulators - paraffin, rosin, probably also celluloid and polystyrene. The first two can be melted and poured, besides the paraffin dissolves in gasoline, you need to be careful with rosin, because at a temperature slightly above the melting point, it starts to char and the insulation properties deteriorate, and both must be very clean so that they insulate well and withstand high temperatures. tension; I think celluloid dissolves in acetone, polystyrene also dissolves in something. Probably the easiest way is with paraffin, but it must be clean, bought in a chemical store, and not from a paraffin candle (such a candle usually has impurities and it insulates much worse). But "hochspannung gel" should be better. One of the best insulators, but generally impractical, is sulfur (unfortunately it is corrosive to metals so cannot be used). Teflon is good, but you won't flood the system with Teflon.
As for PNW-57 - make sure that it will not be overloaded.
Voltage measurement: basically you need a resistor (or resistor array) that can withstand this voltage. Up to 3kV, I used a few (I don't remember, I think 6) 10M resistors (probably 1/2 W or 1W) connected in series. But be careful with their placement on the board - it must provide good insulation. And you need to determine how many resistors you need not to exceed the permissible voltage (for resistors there is some acceptable voltage, with such a high resistance, the power is "fictitious" - supposedly 3kV at 10M will give 0.9W power, but the 10M 1W resistor cannot be connected to 3kV , it will destroy it - the permissible voltage is specified in the catalog note). And you need to know which output the high voltage is at - it's not like the "voltage" and "ground" outputs are equivalent, "ground" may not have good insulation. Measure the current flowing through these resistors and calculate the voltage.
Note: it is better not to connect the resistors to a working power supply, but before turning it on - charging the mounting capacities may break the resistor. Can I do so that there are (small) capacitances parallel to the resistors - so that when the voltage suddenly changes, it is transferred evenly to all resistors? But then this voltage change will give a current impulse to the meter, you need to protect it somehow, e.g. by connecting a capacitor (not so small) to it between its terminals.
In general, the technique of 15kV voltages is difficult - some conductor may discharge into the environment (into the air) because it is too thin.
Do you have information about the current you need to power this lamp? The current below 1mA should not be felt (it does not matter from what voltage), but touching the capacitor charged to 15kV you will probably feel a lot already at a capacity of e.g. 100pF.