Hi Aaron,
I spent some more time at it today and now I can see the same. In your video whenever the radiant event occured, C2 would drop from 1000V to around 500V. Since C2 is positive on the side connected to the coil and the grid, that means that during the radiant event electrons flowed to that side. For that to happen the electrons must have come from the HV rod. That would explain why at lower voltage, electrons arc from the grid to the HV rod, the grid has sharper edges than the rod and so breakdown happens relatively easily from grid to rod. But to break down from rod to grid takes a much higher voltage, and so happens only after a build-up period. Of course, the breakdown voltage would be lower than normal because of all the ionization going on in the tube.

My point, though, is that a Van de Graaff machine is constantly sucking electrons from the HV rod so the rod will never get to break down. With a pulsing power supply, when the pulse is off, if the HV rod is at its breakdown voltage at the time then it can breakdown and send electrons to the grid.

I still want to stick with a high voltage, low current power supply, so I'll probably introduce a spark gap between the Van de Graaff machine and C1 to get the pulsing. I'll put a diode in there too to prevent oscillations.
-Steve