Consider a finite probability space with $\sigma-$algebra $A$ and its generating partition $P$. Let $X,Y \in L^2(A,P)$, then it holds: $$E[X| A]= \sum_{p \in P} 1_p E[X|p] $$.
Proof: $$E\left[Y \sum_{p \in P} 1_p E[X|p]\right]= \sum_{p \in P}E[Y 1_p E[X|p]] = \sum_{p \in P}E[ Y1_p X] = E[YX]$$.
I do not know why I need that extra random variable $Y$. Would it not still work without it?
Without the $Y$ you are only proving that the two sides of the equation $$E[X| A]= \sum_{p \in P} 1_p E[X|p] $$ have the same expectation. To show that $U=V$ is is not enough to show that $EU=EV$. If we show that $EYU=EYV$ for all bounded randoem variables $Y$ (or just indicator r.v.'s $Y$) we can conclude that $U=V$ a.s..
[Proof: Take $Y=1_{U>V}$ to see that $E(U-V)1_{U>V}=0$. A non-negative r.v. cannot have zero expecation unless it is $0$ a.s., so we must have $P(U>V)=0$. Similarly, $P(U<V)=0$, so $U=V$ a.s.].