This circuit is to make the FT turn off faster than it turns on.
FETs tend to have a relatively large requirement for charge into/out of the gate for switching. Although they can stay on with no current, to switch them quickly, they need a large current during the transition.
This circuit turns the FET on with whatever current the driver IC can source, flowing through diode D1.
However, it turns it off with a transistor used as an emitter follower, which can be a much larger current. If it only had to turn off quickly, then a resistor in the place of D1 would be OK. However, it needs to turn on fairly quickly, D1 allows a large forward current to flow, and still allows the driver to bias the transistor correctly to turn the FET off.
Losses may well be reduced if two series FETs use this circuit. There is a problem called 'shoot through', which happens when two series FETs are switched at the same time. If one turns on before the other has turned off, they are both on together briefly, and a large current can shoot through. This modification to the driver ensures that the off-going FET switches faster than the on-going FET, avoiding shoot through.