There should be some current-limiting resistor in series with the base. If you apply 3.3V the red LED and the transistor base will see a large and poorly defined (and possibly destructive) current.
So let's put a 1K resistor in series with the base.
Case 1: Input is 0V. The transistor T1 is 'off', R1 supplies base current to T2 and the green LED has about 2-3V across it. The green LED current is poorly defined by will be limited by R8 and the transistor voltage drop.
Case 2: Input is 3.3V. T1 is 'on', the emitter will be at about 1.8 or 2V and the collector at about 0.1V higher so the base of T2 will be at about 1.9 to 2.1V. The emitter of T2 will be about 600mV less, or about 1.3 to 1.5V. That's not enough voltage to turn a green LED on (they require around 2 to 3V depending on the type, and the current).
Let's do a simulation with the available Circuitlab models:
simulate this circuit – Schematic created using CircuitLab
As you can see, it works, however the green LED is getting quite a bit of current- about 25mA (a result of the Circuitlab green LED model) and the red one a more reasonable 15mA.
This is not a great circuit (due to the poor control of LED current). It does allow a dual red/green led with common cathode to be used, but there are better ways.
Note that if you were to swap the red and green LEDs, the one driven by T2 (now the red one) may not completely turn off because red LEDs require less voltage to reach a given current and brightness. In such a case, the difference between forward voltage of the different colors works against you rather than for you.
Also note that T1 is likely saturated when the red LED is on, however T2 is never saturated even when the green LED is on, because it's an emitter-follower. That means it will have 0.6 or 0.7 volts across it (from collector to emitter) when it is on, whereas Q1 may have only tens of mV or 100mV across it if it's getting enough base current.
Here is one way to do it so that either LED can be either color and both transistors are saturated:
simulate this circuit
The input must be driven high or low. If it is left open then both LEDs will be somewhat 'on'.