Let $\mathcal{C}$ be a small filtered category and $F:\mathcal{C}\rightarrow\textbf{Ab}$ be a functor. Let $U:\textbf{Ab}\rightarrow\textbf{Set}$ be the forgetful functor. Then the filtered colimit $$\left(\coprod_{C\in\mathcal{C}}U(F(C))/\sim,(s_C)_{C\in\mathcal{C}}\right)$$ of the functor $U\circ F$ exists. We want to construct an abelian group structure on the set $\coprod_{C\in\mathcal{C}}U(F(C))/\sim$.
Let $C,C'\in\mathcal{C}$ and $x\in F(C)$, $y\in F(C')$. Since $\mathcal{C}$ is filtered, there exists an object $C''\in\mathcal{C}$ together with morphisms $f:C\rightarrow C''$ and $g:C'\rightarrow C''$ such that $s_{C''}(F(f)(x))=s_C(x)$ and $s_{C''}(F(g)(y))=s_{C'}(y)$. Let $$s_C(x)+s_{C'}(y):=s_{C''}(F(f)(x)+F(g)(y)).$$
I want to show that this operation does not depend on the choices made. Is it enough to pick elements in $x'\in F(C)$ and $y'\in F(C')$ such that $s_C(x)=s_C(x')$ and $s_{C'}(y)=s_{C'}(y')$, or do I have to pick different $C$'s in $\mathcal{C}$ as well?
Supposing that merely picking $x'\in F(C)$ and $y'\in F(C')$ such that $s_C(x)=s_C(x')$ and $s_{C'}(y)=s_{C'}(y')$ suffices, let us show that $s_C(x)+s_{C'}(y)=s_{C}(x')+s_{C'}(y')$. Once again there exists an object $C'''\in\mathcal{C}$ and morphisms $f':C\rightarrow C'''$ and $g':C'\rightarrow C'''$ such that $s_{C'''}(F(f')(x'))=s_C(x')$ and $s_{C'''}(F(g')(y'))=s_{C'}(y')$. What should be the next step? Any hints?