Electronic – Body Diode of a MOSFET

I think you've got it mixed up. The body diode is a part of the structure of the MOSFET and does not help when switching an inductive load with a single MOSFET.

The zener in the symbol represents the avalanche rating of the MOSFET and is of use when switching an unclamped inductive load. You won't see it conduct unless you exceed the voltage rating \$V_{DS}\$ and get to \$V_{(BR)DSS}\$.

If you are actually getting +/-12V out of the pulse transformer, I don't see any obvious reason why that part would not work. Your negative swing out of the pulse transformer does have to be sufficiently large to turn the control MOSFET on, when loaded with both gate charges (-10V would be good).

You have one mistake in your interpretation. In the schematic the internal diode is pointing from source to drain (the FET is flipped from the figure, not rotated).
When the gate voltage is positive, current will flow from drain to source, which matches the schematic if the internal diode is from source to drain.
Another way to tell would be the arrow on the bottom leg of the FET in the schematic. This corresponds to arrow on the figure (the tiny arrow on the source leg). This means the bottom leg is the source, so the internal diode is going from source to drain.

It matters if you switch the pins in that the FETs won't operate correctly. Instead of being switched on, they will simply flow current continuously through that internal diode. As long as everything downstream of the FET can handle that constant current it will be fine, just won't operate as intended.
So for prototyping it won't hurt anything and you'll just have to flip it around. If you had a production board with it backwards it would be a huge issue though, since your FET would act as a short and depending on your pad layout it might be difficult to flip around.