Sharing the low voltage rail of unknown potentials... How to make Proper Floated Voltages?

I asked a previous question about running circuitry off a high voltage supply with unknown ground potential. The question is here: Generating a 10 volt rail from a High Voltage Supply with unknown Ground Potential

Based on the excellent answers, I realized I may have been making it a bit hard on myself. I have a new question:

Let's assume we have a high voltage source as a large capacitor bank at 200 volts. I have a 12 volt battery I want to use to run an NE555 and a mosfet that is switching the 200 volt circuit.

But let's make it more complicated by assuming I can't guarantee that the 200 volts across the cap is +200 and 0. It might be +100 and -100 (or even +10 and -190), whereas my battery might be +12 and 0 relatively speaking. I need to guarantee that I have 12 volts between the gate and source of mosfet so as to not damage it. The question is: If I simply connect the 12 volt battery right into the circuit, will its low potential automatically match up (Float?) to the low potential on the high voltage circuit, thus guaranteeing a 12 volt difference to between the Gate and Source of the mosfet?

And for a more advanced system: What if the 12 volts isn't even supplied by a battery? What if it's just another capacitor bank?

And for a really complicated system: What if I use the 12 volt and GRND rails from a Switched Mode Power Supply like a homebrew ATX power supply? I tried something like that a long time ago. But I was using rectified power from the 120v 60Hz Mains and filtering it into a capacitor. Then I attempted to share the ground rail with an ATX power supply and I ended up destroying the power supply (smokes and sparks!). I never did find out why, but it might have had something to do with the fact that my SMPS and main circuit voltage were both sharing the mains... thoughts?

But let's make it more complicated by assuming I can't guarantee that the 200 volts across the cap is +200 and 0. It might be +100 and -100 (or even +10 and -190), whereas my battery might be +12 and 0 relatively speaking.

You can pick any node in your circuit you want and declare it to be "ground" for that circuit. So you can pick the more negative terminal of your capacitor and call that ground. Then you know the positive terminal is +200 V (relative to the ground that you defined).

If I simply connect the 12 volt battery right into the circuit, will its low potential automatically match up (Float?) to the low potential on the high voltage circuit, thus guaranteeing a 12 volt difference to between the Gate and Source of the mosfet?

Yes. If you connect a net of an isolated circuit into your circuit, you then make the potential of the two connected wires equal. In this case, you make the voltage of the negative terminal of the 12 V battery 0 V by definition.

And for a more advanced system: What if the 12 volts isn't even supplied by a battery? What if it's just another capacitor bank?

What really matters is Kirchoff's Voltage Law. If you add up all the voltages of the branches around a loop, they must add up to zero. The law doesn't care which node is labelled as ground.

And for a really complicated system: ...

Without a schematic of what you connected, nobody can tell you why that situation caused a problem.

You can use a 9V battery for supplying negative voltages relative to ground as well as positive ones. It just depends on which terminal you connect to ground. Providing your 12V is isolated from other power supplies on the system it doesn't matter - it will "float" itself to whatever point you wish to call ground.

I can't comment on the ATX power experiment other than to say I believe what happened is that you either shorted something out or anticipated a connection was floating (isolated) when it wasn't.