How to prove that for two morphisms $f,\psi: F \rightarrow G$ of sheaves of sets on a topological space $X$ if $f = \psi$ if and only if $f, \psi$ induce the same maps on stalks. Deduce that a morphism of sheaves of sets on $X$ is an epimorphism if and only if it is onto on stalks.
I know that we have a theorem about $f$ morphism of sheaves on a topological space that $f$ is isomorphism iff the induced map on the stalk $f_p$ is an isomorphism for every $p\in X$.
You don't have to use any surjectivity hypothesis, since this is a general fact and it is true in general for any morphism of sheaves.
A morphism of presheaves $\phi:F\to G$ is a a collection of maps $\phi_U:F(U)\to G(U)$ for any open set $U\subseteq X$ such that every time you have $V\subseteq U$ and the restricion maps $\rho_{UV}:F(U)\to F(V)$ in $F$ and $\tilde{\rho}_{UV}:G(U)\to G(V)$ in $G$, then $\tilde{\rho}_{UV}\circ\phi_U=\phi_V\circ\rho_{UV}$.
If $f=\psi:F\to G$, then $(\implies)$ is obvious: two equal morphisms induce exactly the same morhpisms on stalks.
For the reverse implication, if $f_x=\psi_x$, you have to prove $f=\psi$. To verify that those two maps are the same, you have to check that they are "the same collection of maps", an so that $f_U=\psi_U:F(U)\to G(U)$ for any open set $U\subseteq X$. So, take $U\subseteq X$ an open subset and a section $s\in F(U)$. If $f_U(s)=\psi_U(s)$ we are done. Take $x\in U$ and denote $s_x=[s,U]\in F_x$ the germ of $s$ in $x$. By hypothesis $f_x(s_x)=\psi_x(s_x)\in G_x$ and so the germs of $f_U(s)$ and $\phi_U(s)$ in $x$ are equal. This is equivalent to say that exists $V\subseteq U$ an open neighborhood of $x$ such that $f_U(s)|_V=\psi_U(s)|_V$. This is true in a neighborhood of each point of $U$ and you can conclude $f_U(s)=\psi_U(s)$, because $G$ is a sheaf and "if two sections are locally the same then they are equal".
Note that in the proof we have used only the properties of sheaf of $G$ and you can conclude that this is true also if $F$ is a presheaf.
A morphism of sheaves $\phi:F\to G$ is said surjective if for each $x\in X$ there exists an open neighborhood $U\ni x$ such that $\phi_U:F(U)\to G(U)$ is surjective. It is possible that you have seen another definition, but this should be equivalent. Notice that this definition is WEAKER than asking $\phi_U:F(U)\to G(U)$ surjective for each open set $U$. However this stronger condition is equivalent to the definition if $\phi$ is injective...
You want $\phi:F\to G$ surjective $\Longleftrightarrow$ $\phi_x:F_x\to G_x$ surjective for each $x\in X$. Note that the condition $\phi_x:F_x\to G_x$ surjective is equivalent to say that for each germ $[t,U_1]\in G_x$ there exists a germ $[s,U_2]\in F_x$ such that $\phi_x([s,U_2])=[t,U_1]$ and as you know those two germs are the same if and only if exists an open neighborhood $V\subseteq U_1\cap U_2$ of $x$ such that $\phi(s)|_V=t|_V$ and this is equivalent to say that $\phi$ is surjective.