Let $X$ be a topological space, and let $\alpha : \mathscr{F} \to \mathscr{G}$ be a morphism of sheaves on $X$. Let $j : \{ x \} \hookrightarrow X$ be the inclusion of a point into $X$. Then, by definition, the stalk of $\mathscr{F}$ at $x$ is just the inverse image sheaf $j^* \mathscr{F}$. But we know $j^* \dashv j_*$ (i.e. the inverse image functor $j^*$ is left adjoint to the direct image functor $j_*$), and left adjoints preserve colimits, hence if $\alpha$ is epic, so is $\alpha_x : \mathscr{F}_x \to \mathscr{G}_x$.
Conversely, suppose $\alpha_x : \mathscr{F}_x \to \mathscr{G}_x$ is surjective for every $x$ in $X$. Let $\beta, \gamma : \mathscr{G} \to \mathscr{H}$ be two further morphisms of sheaves, and suppose $\beta \circ \alpha = \gamma \circ \alpha$. We need to show $\beta = \gamma$. Certainly, we have $\beta_x \circ \alpha_x = \gamma_x \circ \alpha_x$, and $\alpha_x$ is epic by hypothesis, so we have $\beta_x = \gamma_x$ for all $x$ in $X$. Let $s \in \mathscr{G}(U)$, and consider $\beta_U(s)$ and $\gamma_U(s)$. Since $\beta$ and $\gamma$ agree on germs, at each $x$ in $U$ there is an open neighbourhood $V$ on which $\beta_V(s|_V) = \gamma_V(s|_V)$, and so by the unique collation property, we must have $\beta_U(s) = \gamma_U(s)$, and therefore we indeed have $\beta = \gamma$.
For your first question, yes, that's exactly what it means to say that the kernel sheaf is 0. For your second question, have you studied complex analysis? The identity theorem states that if two holomorphic functions agree on an open subset, then they are identical wherever they are both defined. Thus it makes sense to identify holomorphic functions that agree on some open subset and only consider equivalence classes, aka germs. This is exactly the equivalence relation defining the stalk of a sheaf.
The stalk of a sheaf is defined as the direct limit of the sheaf with its restriction maps:
$$
\mathcal{F}_p = \varinjlim_{U \ni p} \mathcal{F}(U)
$$
taken over all open sets $U \subseteq X$ containing $p$. Given a morphism $\varphi$ of sheaves, the universal property of the direct limit yields an induced map on the stalks. More concretely, we can choose a representative $(s,U)$ for a germ $s|_p$. Since the morphism commutes with the restriction maps, then
\begin{align*}
\varphi_p(s|_p) &= \varphi_p(\operatorname{res}^U_p(s)) = \operatorname{res}^U_p(\varphi_U(s)) \, .
\end{align*}
(Note that the first restriction map is for the sheaf $\mathcal{F}$, $\operatorname{res}^U_p: \mathcal{F}(U) \to \mathcal{F}_p$, while the second is for the sheaf $\mathcal{G}$, $\operatorname{res}^U_p: \mathcal{G}(U) \to \mathcal{G}_p$.)
So to compute the action of the stalk map $\varphi_p$ on a germ, we just compute the action of the morphism on a representative for the germ, and then restrict to the stalk.
EDIT: Here's a solution to the problem:
Assume $\varphi_p$ is injective for all $p$. Suppose $U$ is an open set and $s \in \ker(\varphi_U)$, so $\varphi_U(s) = 0$. Given $p \in U$, then
$$
\varphi_p(s|_p) = \varphi_p(\text{res}^U_p(s)) = \text{res}^U_p(\varphi_U(s)) = \text{res}^U_p(0) = 0 \, .
$$
Since $\varphi_p$ is injective, then $s|_p = 0$, so there exists an open set $V_p \subseteq U$ with $p \in V_p$ such that $\text{res}^U_{V_p}(s) = 0$. Note that the collection $\{V_p\}_{p \in U}$ forms an open cover of $U$. Since $\mathcal{F}$ is a sheaf and $\text{res}^U_{V_p}(s) = 0$ for all $p \in U$, then $s = 0$ by local determination (the first sheaf axiom). Thus $\ker(\varphi)(U) = \ker(\varphi_U) = 0$ for all $U$.
Conversely, assume $\ker(\varphi_U) = 0$ for all $U$. Given $s|_p \in \ker(\varphi_p)$, choose a representative $(s,U)$ for $s|_p$. Then
$$
0 = \varphi_p(s|_p) = \varphi_p(\text{res}^U_p(s)) = \text{res}^U_p(\varphi_U(s)) \, .
$$
Then there exists an open set $V \subseteq U$ containing $p$ such that
\begin{align*}
0 = \operatorname{res}^U_V(\varphi_U(s)) = \varphi_V(\operatorname{res}^U_V(s)) \, .
\end{align*}
Since $\varphi_V$ is injective, then $\operatorname{res}^U_V(s) = 0$, so
$$
s|_p = \operatorname{res}^U_p(s) = \operatorname{res}^V_p(\operatorname{res}^U_V(s)) = \operatorname{res}^V_p(0) = 0 \, .
$$
Best Answer
As indicated by the KReiser hint, you can do as follows:
We see the following diagram :
The first and second rows are exact , due to sheaf property, the second square commute do to compactible with restriction. and the universal property of the kernel gives the morphism we want.