I should have realised that the circuit would be a little bit 'fringe', being posted on here
OK let's dissect this one stage at a time.
Firstly it's a 3 phase generator, I've built one of these although in a slightly different form, they work fine.
Secondly you need to have a close look at the region around the base of the transistor, where the capacitor joins the transistor. The voltage here goes to +0.7v (when the transistor is on), and then down to a good few negative volts when the transistor is off. The combination of the capacitor and the resistors mean that the voltage it gets down to depends on the oscillation frequency, I'd imagine somewhere around -5V * or so.
Now look at where the other probe of our voltmeter is going: to the switched side of a lamp that can go up to 15V when it's off.
Add the two together, and voila, 20V, easy as pie.
The problem here is that the person who's doing the measuring has confused voltage with power. Once you start putting a load on the output, the output voltage will drop, and the frequency will change as well. There is no 'free energy' in this circuit, all that's being measured is an oscillator with a very inefficient voltage doubler.
A better circuit that will do the same thing is here:
Think about this: You can buy "inverters" from any car or camping place, that take 12V in, and put out 120 or 240V. Is this free energy? Of course not. You need a hefty amount of current on the 12V side, and you don't get much current on the 240V side. You end up putting in about 30% more power into the 12V side than you get out in the 240V side.
* That minus 5V is probably what's burning out the transistors -- they really really DON'T like having negative voltage between the base and emitter. To fix this you would add a resistor in series with each capacitor, as well as a small diode to make sure that voltage didn't go too negative. This would, unfortunately, lower the voltage that you read on the 'output' multimeter.