With normal loads, the power transistor is in saturation and so the regulated +12V line is directly available (minus only the Vcesat of the PNP transistor). If the output is accidentally shorted, the npn transistor turns off and removes the base drive of this transistor.
This circuit is intended for protection from short circuits. It might be possible to apply a load that brings the series transistor out of saturation without removing base drive to it: such a situation will cause heating and might cause it to fail short-circuit. Those who wish to draw significant current from this circuit will be well advised to check that the transistor is indeed in saturation with their intended load.
That said, this is a simple way to draw a few tens of milliamps at 12 V from your PC without danger to either load or the PC.
Step 1: The circuit
A small signal NPN transistor is fed with the output voltage through two resistors connected as a potential divider. Should the output voltage become less than a certain value, this transistor will turn off.
The collector of this transistor supplies the base current of the power PNP transistor. So the two transistors will always switch on or off together.
The power transistor (BD140) is supplied with a certain current via the green LED and 470 ohm resistor. It has a certain gain, around 50 or so, and its collector can supply only the gain times base current before it drops voltage (comes out of saturation).
So the maximum current draw can be changed by appropriately dimensioning the base resistor of the PNP power transistor.
The voltage divider at the output determines the voltage at which the current supplied by the transistor drops to zero.
The 470 ohm resistor in parallel with the PNP transistor supplies a certain amount of current so that the circuit will switch on when first powered on.