Entropy of sum is sum of entropies Having $X$ and $Y$ discrete random variables above finite set. Z is defined as $Z=X+Y$ when does the following happen:

$$H(Z)=H(X)+H(Y)$$
That is not true in general. 

$$H(X,Z)=H(Z)+H(X \mid Z) = H(X)+H(Z \mid X) $$

But $H(Z \mid X)=H(X+Y \mid X)=H(Y \mid X)$. 

Further, if we assume $X,Y$ are independent, then $H(Y \mid X)=H(Y)$ and 

$$ H(Z)=H(X)+H(Y)-H(X \mid Z)$$ 

We still need $H(X \mid Z)=0$ (knowing the sum lets me know the summands). This is only true in the very special case in which $x_1+y_1 \ne x_2+y_2$ for any set of values with positive probability.
If both $X$ and $Y$ comes from the same set, then it's only true in the trivial case of a single element support.
I am afraid H(Z|X)  is not equal to H(Y|X), if X and Y are independent and Z = X+Y, because say X and Y are binary and independent, then X and Y say nothing about each other, but Z = 0 => X = 0, and Z = 2 => X = 1. So H(X|Z) is not equal to H(X|Y)=H(Y).