Here is a proof question: For two random variables $X$ and $Y$, we can define $E(X|Y)$ to be the function of $Y$ that satisfies $$E(Xg(X)) = E(E(X|Y)g(Y))$$ for any function $g$. Using this definition show that $E(X_1 + X_2|Y) = E(X_1|Y) + E(X_2|Y)$.
So what I did was I plugged in to $X = X_1 + X_2$. $$E(E(X1 + X2)|Y)g(Y)) = E(X1g(Y)) + E(X2g(Y)) = E(E(X1|y)g(Y) + E(X2|Y)g(Y)) = E(g(Y) [E(X1|Y) + E(X2|Y)]$$
am I on the right track? what do I do after that?
Yes, you are on the right track. As a final touch you should appeal to the uniqueness of the conditional expectation to show that $E((X_1 + X_2)\mid Y) = E(X_1\mid Y) + E(X_2\mid Y).$ Remember that $E(X\mid Y)$ is the almost surely unique $\sigma(Y)$-measurable random variable that satisfies $E(Xg(Y)) = E(E(X\mid Y)g(Y)).$