Let $X$ be a topological space and associate to each open subset $U \subset X$ a set $S(U)$ in such a way that whenever $V \subset U$ is another open subset the so called restriction maps $\rho_V^U \colon S(U) \to S(V)$ satisfy the condition $$\rho_W^V \circ \rho_V^U = \rho_V^U \qquad \rho_U^U = \text{id}_U$$ for all open $W \subset V \subset U$. The collection $S = \{S(U), \rho_V^U\}$ is then called a presheaf on $X$. Now given a presheaf $S = \{S(U), \rho_V^U\}$ we may consider the disjoint union $$\mathcal{U}_x = \bigsqcup\limits_{U \ni x} S(U)$$ where the union is taken over all open subsets $U$ which contain $x \in X$. For $f \in S(U)$ and $g \in S(V)$ define the relation $$f \sim g \iff \exists x \in W \subset U \cap V \colon \rho_W^U(f) = \rho_W^V(g)$$ Quite evidently this relation is reflexive and symmetric. However, why does it satisfy transitivity? If $h \in S(N)$ such that $f \sim g$ and $g \sim h$, then by assumption there are open neighborhoods $W_1, W_2 \subset X$ of $x$ such that $\rho_{W_1}^U(f) = \rho_{W_1}^V(g)$ and $\rho_{W_2}^V(g) = \rho_{W_2}^N(h)$. Setting $W = W_1 \cap W_2$, I guess the idea would be $$\rho_W^U(f) = \rho_W^{W_1} \circ \rho_{W_1}^U(f) = \rho_W^{W_1}(\rho_{W_2}^V(g)) \underset{\text{Is this valid?}}{=} \rho_W^{W_2} \circ \rho_{W_2}^{V}(g) = \rho_W^{W_2} \circ \rho_{W_2}^{N}(h) = \rho_W^N(h)$$ I don't quite see how one could justify the transition from $W_1$ to $W_2$ in the upper index of $\rho$. I either probably overlook something completely trivial, or I am just mistaken (or both), so I guess the question is:
How to properly show transitivity of this relation?
You missed a single step: $$\begin{align}\rho^U_W(f)&=\rho^{W_1}_W(\rho^U_{W_1}(f))&\text{composition of restrictions}\\ &=\rho^{W_1}_W(\rho^V_{W_1}(g))&\text{assumption $f\sim g$}\\ &\color{red}{=\rho^{V}_W(g)}&\text{composition of restrictions}\\ &=\rho^{W_2}_W(\rho^V_{W_2}(g))&\text{composition of restrictions}\\ &=\rho^{W_2}_W(\rho^N_{W_2}(h))&\text{assumption $g\sim h$}\\ &=\rho^N_{W}(h)&\text{composition of restrictions}\\ \end{align}$$
In fact, one clearly has that sections that are equal after restriction are still equal after stronger restriction.