The diode will only see the 2 V forward, so the current will flow. The diode is directly connected to the 2 V generator so everything outside doesn't matter (excepted short-circuit and so on of course), if you can measure 2 V at the generator then there is also 2 V at the diode.
But you'll not have 2 V at this point, you'll have the diode drop (0.6 to 0.7 V for silicon and 0.3 for schottky diodes for example) or a fried diode if the current is too high...
EDIT (Since the OP has edited his question my answer just above is no longer valid):
The diode will only see 2 V in reverse (or -2 V in the conventional reading direction of a schematic), so the current will not flow. The diode is directly connected to the 2 V generator so everything outside doesn't matter, if you can measure 2 V at the generator then there is also 2 V at the diode.
So you'll have 9 V accross the whole thing, -2 V accross the diode and +9 + -2 = 7 V accross R1.
EDIT² (answer to the secondary question: Where will the current flow when V1 = 9 V and then V1 = 0 V?)
Since the diode isn't conducting there is only one current path and it will be from V1, through V2, then trough R1, then to GND when V1 = 9 V.
When V1 = 0 V (connected to GND in other words) the path will be the same but the current will flow in the reverse direction (but the diode will still be reverse biased and not conducting, don't forget the diode is still directly connected to the generator V2 so everything outside doesn't matter).
From a different point of view: if you think that V2 is like a battery, when V1 = 9 V you recharge it through R1, and when V1 = 0 V then you discharge it directly through R1.
EDIT³ (new schematic and new secondary question: Where will the current flow when Q1 is closed and then opened?)
With this new schematic when Q1 is closed (let's assume it's a perfect transistor) the path and current direction will be the same as in my edit² when V1 = 9 V.
When Q1 is opened there is no path where the current can flow so there is simply no current flowing.