Hi CW,
It is a classic problem with push-pull transistors to make sure that both transistors do not conduct at the same time- bad for the transistors, bad for the transformer, bad for power efficiency, and bad for output waveform purity.
The upper two NMOSFETs operate in parallel as do the two lower NMOSFET pair, so each parallel pair can be considered to be one bigger NMOSFET
As the other members have implied, and as you suspect, the diode in parallel with the 10 Ohm resistor turns the NMOSFETS on relatively slowly and off fast.
But, in order to understand how this works, you need to know one vital bit of information about MOSFETs: they have a very high gate/drain capacitance.
In order to charge and discharge the gate capacitance fast, a gate driver chips are used, as is the case in your circuit, Gate drivers have a high source and sink current capability, typically 1.5A.
Now to the core point; the the gate capacitance is charged up via the 10 ohm resistor (turn on) and is discharged via the 10 Ohm resistor in parallel with the forward biased diode, which could have a forward effective resistance of 0.5 Ohm.
So after all that, the you can see that the turn on gate time constant is 10 ohms * CGD (capacitance, gate/drain) and, roughly speaking, the turn off gate time constant is 0.5 ohms * CGD, ie 20 times faster.
That is not the whole story about gate driving, but I hope it answers your question.
spec