Consider the operation of the circuit.
When the transistor is on current is flowing in the coil from top to bottom as the circuit is drawn we now switch the transistor off. The current in the coil still wants to flow.
For the circuit on the left this current can now flow back to Vcc via the diode the voltage across the coil has reversed direction and is limited by the diode the current can decay to zero safely.
For the circuit on the right the diode does not help. The current flowing in the coil will force the voltage on the collector to rise to the point where the transistor (or possibly the diode) breaks down and starts to conduct. At this point the current can start to decay in the coil but the energy in the broke down transistor (or less likely diode) will be excessive and may well result in the transistors death. Note a zener diode here will work because you allow the voltage on the coil to reverse so the current can decay to zero while limiting the voltage across the transistor to a safe value.
It should be noted the allowing the voltage across the coil to reverse to an higher voltage means the current can decay more quickly which is why you sometimes see a zener in the right hand circuit or more than one diode in series in the left hand one.
The voltage spike created by the leakage inductance of your transformer is breaking down your MOSFET each time it switches off. You need to limit this voltage somehow; the usual technique is to put an R-C snubber across the source and drain of the MOSFET, sized so that the capacitor captures the energy of the
spike before the voltage rises too high.
The values required depend very much on the characteristics of your specific transformer, so you'll have to do some experimentation to determine them.
One way to start is to find the value of the peak primary current in the transformer; the resistor should be sized so that this current, multiplied by the resistance, gives a peak voltage that's comfortably less than the rating of the MOSFET.
At the moment the MOSFET switches off, the drain voltage initially spikes high because of the leakage inductance of the transformer, but then it settles down to a voltage that's proportional to the secondary voltage (by the turns ratio of the transformer). The capacitor needs to be sized so that it charges to that voltage level in a time that's somewhat longer than the duration of the spike. This depends on both the inductance value and the resistor value. Be conservative at first (i.e., use an over-large value) and then fine-tune it (for better efficiency) once you have the circuit working.
Best Answer
No. If the inductor were perfect, and the switching of the FET instantaneous, then the inductor would generate an infinite voltage when you turned off the switch. Even in our imperfect world, putting the diode in series with the FET source will just burn out both devices while protecting neither.
Inductors do not want to change current instantaneously. You must give that current someplace to go when you turn off the FET. There are various ways to arrange catch diodes, but you have to answer the question "where will the current go if it WILL go someplace?"