Let $\mathcal F$ be a sheaf of a topological space $X$ and $D_{\mathcal F}(U)=\prod_{U\ni x}\mathcal F_x$.
I don't understand this morphism $\varphi_{\mathcal F}:\mathcal F\to D_{\mathcal F}$, when we apply in the open subset $U$ we have $({\varphi_{\mathcal F}})_U:\mathcal F(U)\to D_{\mathcal F}(U),\ s\mapsto (s_x)_{x\in U}$ but what exactly means $s\mapsto (s_x)_{x\in U}$?
Can I say that $s\mapsto (\overline{(s,U)},\overline{(s,U)},\ldots)$? since $s\in \mathcal F(U)$, we have $s_x=\overline{(s,U)}$ for every $x\in U$.
I think maybe I didn't fully understand the meaning of $s_x$.
I really need help.
Thanks a lot.
An element in a product over a set of indices $U$ is an arrow from the set of indices to the coproduct: so your $D_\mathcal F(U)$ is $$\prod_{x\in U}\mathcal F_x=\textrm{Maps }(U,\coprod_{x\in U}\mathcal F_x).$$
Thus, concretely, your morphism $\phi_\mathcal F:\mathcal F\to D_\mathcal F$ on an open subset $U\subset X$ sends a section $s\in\mathcal F(U)$ to the map $U\to \coprod_{x\in U}\mathcal F_x$ defined by $x\mapsto s_x$, where $s_x\in\mathcal F_x$ is the germ of $s$ at $x$, that you correctly described as an equivalence class.