$\require{AMScd}$ Let $\phi:\mathscr{F\to G}$ be a morphism of sheaves on $X$, let $\mathscr F_P$ be a stalk of $\mathscr F$ at $P\in X$, and let the stalk map $\phi_P:\mathscr F_P\to\mathscr G_P$ take the class containing $(U,f)$ to the class containing $(U,[\phi(U)(f)])$. We aim to show that this defines a function, by showing that it is well-defined. The requirement is to show that if $(U,f)\sim(V,g)$ then $(U,\phi(U)(f))\sim(V,\phi(V)(g))$. The assumption implies that there is an open set $W\subseteq U\cap V$ such that $f|_W=g|_W$. From the definition of morphism, we have commutative diagrams $$\begin{CD} \mathscr F(U) @>\phi(U)>> \mathscr G(U)\\ @VV\rho_{UW}V @VV\rho'_{UW}V\\ \mathscr F(W) @>\phi(W)>> \mathscr G(W) \end{CD}\qquad\qquad \begin{CD} \mathscr F(V) @>\phi(V)>> \mathscr G(V)\\ @VV\rho_{VW}V @VV\rho'_{VW}V\\ \mathscr F(W) @>\phi(W)>> \mathscr G(W) \end{CD}$$ which imply $$\rho'_{UW}\circ\phi(U)=\phi(W)\circ\rho_{UW}\qquad\qquad\rho'_{VW}\circ\phi(V)=\phi(W)\circ\rho_{VW}.$$ This gives $$\phi(U)(f)|'_W=\rho'_{UW}(\phi(U)(f))=\phi(W)(\rho_{UW}(f))=\phi(W)(g|_W)$$ and $$\phi(V)(g)|'_W=\rho'_{VW}(\phi(V)(g))=\phi(W)(\rho_{VW}(g))=\phi(W)(g|_W)$$ so that $\phi(U)(f)$ agrees with $\phi(V)(g)$ on $W$ in the sheaf $\mathscr G$, and hence the well-definedness is verified.
Is this rather ugly calculation the best way of showing this simple result? More generally, do most sheaf-related problems ultimately reduce to such computations without sufficient abstraction?