I've found that without the battery V(out)should be 4V, so is this cancelled by the battery? How does the battery add to current in this case?
Superposition Theorem: The total current in any part of a linear circuit equals the algebraic sum of the currents produced by each source separately.
Since you don't have any current source just replace once at a time one voltage source with a short circuit, and leave the other intact. For example, if 10V is shorted first then the voltage across R2 would be 4V, hence a current of 4V/4koms = 1mA would flow through the resistor R2. Now assume 4V is shorted and 10V is on, then R2 would be shorted too and no current would flow through it. Hence the total current flowing through R2 becomes Itot = 1mA + 0A = 1mA.