As it is possible to do the mod, there are going to be a few issues.
Where the OP calls for adding two 1N4148 diodes in series with CR3,(see edit note) on the VTX that will be CR4.(see edit note)
Solder them in series up & off of the board.
It appears that Peavey was trying something, as the Bandit 65 & the VTX, while being identical on the top half of the feedback loop, are markedly different on the output half.
The Bandit mod has another 1N4148 added in series to CR5. No biggy. Bring one end of each diode up off of the board & tie them in series.
Now the VTX already has two diodes each & they are tied together in the middle.
So one of the diodes, CR5 or CR7 has to be removed.
Additionally, the existing diode has to jump across that 'tied together' trace.
EDIT: It appears that the OP had his labels wrong.
According to his attached layout, the top half diode to change is CR4.
On the VTX, that will be CR3.
(which makes more sense , if asymmetrical clipping is the desired outcome.)
Where the OP calls for adding two 1N4148 diodes in series with CR3,(see edit note) on the VTX that will be CR4.(see edit note)
Solder them in series up & off of the board.
It appears that Peavey was trying something, as the Bandit 65 & the VTX, while being identical on the top half of the feedback loop, are markedly different on the output half.
The Bandit mod has another 1N4148 added in series to CR5. No biggy. Bring one end of each diode up off of the board & tie them in series.
Now the VTX already has two diodes each & they are tied together in the middle.
So one of the diodes, CR5 or CR7 has to be removed.
Additionally, the existing diode has to jump across that 'tied together' trace.
EDIT: It appears that the OP had his labels wrong.
According to his attached layout, the top half diode to change is CR4.
On the VTX, that will be CR3.
(which makes more sense , if asymmetrical clipping is the desired outcome.)
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