Why does this circuit speedup the switching time and reduce losses?
If you look at the full schematic, it is easily apparent why. The MC34063 is configured in the circuit in such a way that it drives the MOSFET gate with an open emitter output: The chip can drive the gate high by turning on the NPN transistor between the pins 1 (switch collector) and 2 (switch emitter), but the chip itself has no way of pulling the gate low.  

  

Without the external transistor the only thing discharging the charge stored in the gate capacitance would be the 470 ohm resistor from the gate to ground. This is a slow process, since the resistor can only pass a few tens of milliamps. Selecting a lower resistance would decrease the switch-off time, but the resistor would also conduct more current when the gate is being driven high, wasting energy and heating up in the process.  

The extra transistor Q2 solves this by pulling the gate down immediately after the MC34063 ceases to pull the gate high: When the transistor inside the chip switches off, the resistor pulls the base of Q2 low. Since Q2 is a PNP transistor, it starts to conduct when current is pulled from its base, quickly discharging the MOSFET gate charge to ground. When the chip then turns its output transistor back on, both the collector and base of Q2 are pulled high, preventing base current flow and keeping it switched off.  

The diode D2 blocks the flow of current from the MOSFET gate to the base of Q2, allowing the Q2 base voltage to fall below the Q2 emitter voltage. Without D2 Q2 would be unable to turn on.