Voltage and current in this schematic
(In ideal conditions)
You're correct in your understanding of the first case. When you close the switch, you short one end of the resistor to ground, so there's a 12V drop across it and I=0.02A.

In your second case, with the switch closed, you are creating a voltage divider so the point between the resistors is at 6V.

If you don't have a load connected along Vout (between the black dot and ground), then I=0 there since there isn't a voltage drop along the horizontal connection.

Measuring along the path connecting the resistors/switch, you'll see your 0.01A, and since there is no other path for the current to take, both resistors will see the full 0.01A.