The motor uses its 600mA max current when accellerating, working hard and stalled.
The BD139 is spec'd with a minimum current gain of 25 with 2V across it. Its saturation voltage is spec'd with a much higher base current so that it is turned on as hard as it can. With 500mA collector current its base current is 50mA when its saturation voltage is spec'd, one-tenth. So with the 600mA motor current the base current should be 60mA for the transistor to be turned on as hard as it can.
You probably don't have a high enough supply voltage for the base resistor to be using an emitter-follower circuit:
1) The loss in the 270 ohm resistor is 270 x 60mA= 16.2V.
2) Even if the supply voltage is higher than the voltage wasted in the base resistors, the Vbe of the BD139 is more than 1V maximum with a collector current of 600mA. So the total loss is 17.2V max.
Using a common-emitter NPN transistor with the motor at its collector, the base voltage could be a little higher than 1V with a collector current of 600mA and a base current of 60mA. Then a suitable base reistor will work from the supply and the collector will saturate with a loss of a little more than only 0.5V max.