$Hom(\mathcal{P},\mathcal{P'})$ is additive abelian group for $\mathcal{P},\mathcal{P'}\in pSh(X,Ab)$ This is problem 5.38(ii) of Rotman Homological Algebra. I am not certain about my last step of showing the group is additive.
Let $P,P'\in pSh(X,Ab)$ where $pSh(X,Ab)$ is presheaf category of abelian groups over a topological space $X$. Natural transformation $Hom(\mathcal{P},\mathcal{P'})$ is an additive abelian group. 
It is clear that I have a $0$ map in $Hom(\mathcal{P},\mathcal{P'})$ by $0-sheaf$ is zero object in presheave category with fixed $X$. 
Every element has an inverse by sending $F\in Hom(\mathcal{P},\mathcal{P'})$, $-F\in Hom(\mathcal{P},\mathcal{P'})$ and one can check $F_U-F_U$ for any $U\subset X$ open resulting in $0$ map.
I am not very clear on the additive property. Suppose I take $F,G\in Hom(\mathcal{P},\mathcal{P'})$. Then for any $\mathcal{U}\subset X$, $F_\mathcal{U}:\mathcal{P(U)}\to\mathcal{P'(U)}$. It is clear that I can perform the operation $F_\mathcal{U}+G_\mathcal{U}$ on each $U\subset X$ open. So I can define $(F+G)_\mathcal{U}:=F_\mathcal{U}+G_\mathcal{U}$
Similarly given any $F_1:P_1\to P_2$ and $F_2,F_3:P_2\to P_3$ where $P_i$ are presheaves of abelian groups. $F_i$ are natural transformations. I could check that $F_1\circ(F_2+F_3)=F_1\circ F_2+F_1\circ F_3$ on the abelian group level.
The following are my questions.


*

*Can I define $(F+G)_\mathcal{U}=F_\mathcal{U}+G_\mathcal{U}$ for each $U\subset X$? 

*Was there a easier trick to check $(F_2+F_3)\circ F_1=F_2\circ F_1+ F_3\circ F_1$ without checking the equality on abelian group level?
 A: Your definition for the addition is correct. If you take any $f\in\mathcal P(U)$, then you have
$$(F+G)_U(f)=F_U(f)+G_U(f)\in\mathcal P'(U),$$
since $\mathcal P'(U)$ is an abelian group.
It seems to me that at this point you are done. Why do you want to show that $F_1\circ(F_2+F_3)=F_1\circ F_2+F_1\circ F_3$? First off, there must be a typo here, because if $F_1:P_1\to P_2$ and $F_2,F_3:P_2\to P_3$, then the composition $F_1\circ(F_2+F_3)$ doesn't make sense in the first place.
Edit: I'm going to rewrite $[F_1,F_2,F_3]$ as $[F,F',F'']$ for convenience of notation.
Since by definition, $(F'+F'')_U(g)=F'_U(g)+F''_U(g)$, it really follows without saying much that $(F'+F'')\circ F=F'\circ F+F''\circ F$. If you really wanted to write the details, for $g\in\mathcal P(U)$, we have
\begin{align*}
((F'+F'')\circ F)_U(g)&=(F'+F'')_U(F_U(g))\\
&=F'_U(F_U(g))+F''_U(F_U(g))\\
&=(F'\circ F)_U(g)+(F''\circ F)_U(g)\\
&= (F'\circ F+F''\circ F)_U(g),
\end{align*}
and therefore $((F'+F'')\circ F)_U=(F'\circ F+F''\circ F)_U$ for all $U$.
