L1 is bifilar coil. l2 l3 also.

l1 is connected to plus and minus dc.
the center tap, between the 2 windings of l1 has a mosfet switch.

when switch is closed, a magnetic field is build up in the mutual coupled windings of l1.

when switch is opened, 2 equal but opposite back emf impulses rise.

these pass through the diodes and capacitors into L2.

L2 gets to see opposite polarities on both ends, shorting out in the middle.

L2 now acts as the primary for L3.

l3 ia resonant tuned to the impulses of L2

nope...
capacitors charge up, wont discharge

maybe use tvs diodes to discharge?

A quick analysis of your circuit indicates that when the transistor is inactive the circuit  DC+ - L1a - Diode+B - capacitor1 - L2 - capacitor2 - Diode-B - L1b - DC- gets energized, so when the transistor is active it shorts out the diodes/capacitors/L2 part of the circuit and the current flows from DC+ - L1a - transistor -  L1b - DC-, leaving the charge in the capacitors nowhere to go given that the diodes blocks their (the capacitors) current path through the transistor.

By all means, if you have the time and the energy to it, feel free to point me in the right direction regarding your theory here, in this particular case/circuit, with regards to current paths, charging, discharging, back EMF's, etc etc.