I'll explain based on the circuit on the left.
Theoretically, you could get away using only one transistor, but you wouldn't be making a very good current source, because it would vary greatly with the load you'd be applying to it.
When you analyze transistor circuits, one of the main things you need to remember is that in the active region the voltage drop from base to emitter (output with an arrow) is around 0.6V. What happens in the circuit is the following:
1) the transistor on the lower left (the one with grounded emitter), causes 0.6 drop accross the 27 ohm resistor, and that drop is relatively constant (this gives you the stability), the second transistor (the one with collector connected right to the LED), isolates your current source (the 27 Ohm resistor with 0.6 V accross it), from the load (LED's). It does it thanks to low impedance looking into the emitter, and high impedance looking into the collector.
Another way of looking at this is that the transistor on the left, provides you with a stable biasing for the transistor on the right. You could get away with replacing the transistor on the left with another resistor such that the Voltage at the gate of the first transistor due to the resulting voltage divider, would 1.2 V, and you'd still end up with 0.6 V accross the 27 Ohm resistor.