Given maps $f:X\to Y$ and $g:X\to Y$, a natural way of writing them together in the same diagram would be like this:
$$X{{f\atop {\Large\longrightarrow}}\atop{{\Large\longrightarrow}\atop g}}Y$$
A map $\mathrm{eq}:E\to X$ is an equalizer (Wikipedia) of the maps $f$ and $g$ if it is final in the category of maps to $X$ that equalize $f$ and $g$. Depicting it together with the maps $f$ and $g$, we say that
$$E\xrightarrow{\;\mathrm{eq}\;} X{{f\atop {\Large\longrightarrow}}\atop{{\Large\longrightarrow}\atop g}}Y$$
is an equalizer diagram.
Let $X$ be a topological space, let $F$ be a presheaf on $X$, and let $\{U_i\}_{i\in I}$ be an open cover of $X$. The presheaf $F$ comes with restriction maps $\mathrm{res}_{V}^{U}:F(U)\to F(V)$ for each pair of open sets $U,V$ with $V\subseteq U$. In particular, for each pair of $i,j\in I$, we have restriction maps $$\large\mathrm{res}_{U_i\cap U_j}^{U_i}:F(U_i)\to F(U_i\cap U_j)$$ and $$\large\mathrm{res}_{U_i\cap U_j}^{U_j}:F(U_j)\to F(U_i\cap U_j).$$
By definition, for each $i\in I$, the product (Wikipedia) $\prod_{i\in I}F(U_i)$ comes with a projection map
$$p_i:\prod_{i\in I}F(U_i)\longrightarrow F(U_i).$$ Moreover, $F(U_i)$ comes with a restriction map $\mathrm{res}_{U_i\cap U_j}^{U_i}:F(U_i)\to F(U_i\cap U_j)$ for each $j\in I$. Composing them, we have for each pair of $i,j\in I$, a map $a_{i,j}$, as follows:
$$\underbrace{(\mathrm{res}_{U_i\cap U_j}^{U_i}\circ p_i)}_{\Large a_{i,j}}:\prod_{i\in I}F(U_i)\longrightarrow F(U_i\cap U_j)$$
By the definition of a product, these $a_{i,j}$'s induce a map (let's call it $\alpha$) from $\prod_{i\in I}F(U_i)$ to the product of all of the $F(U_i\cap U_j)$'s together:
$$\alpha:\prod_{i\in I}F(U_i)\longrightarrow \prod_{i,j\in I}F(U_i\cap U_j).$$
Similarly, for each pair of $i,j\in I$, we have a map $b_{i,j}$:
$$\underbrace{(\mathrm{res}_{U_i\cap U_j}^{U_j}\circ p_j)}_{\Large b_{i,j}}:\prod_{i\in I}F(U_i)\longrightarrow F(U_i\cap U_j)$$
which come together to form a single map
$$\beta:\prod_{i\in I}F(U_i)\longrightarrow \prod_{i,j\in I}F(U_i\cap U_j).$$
These maps $\alpha$ and $\beta$ are the top and bottom arrows in the diagram
$$F(U) \longrightarrow \prod_{i\in I} F(U_i) {{{}\atop {\Large\longrightarrow}}\atop{{\Large\longrightarrow}\atop {}}}\prod_{i, j\in I} F(U_i \cap U_j)$$
We say that $F$ is a sheaf (Wikipedia) if, for any open cover $\{U_i\}_{i\in I}$ of our space $X$, this diagram is an equalizer diagram.
Note that $\alpha$ and $\beta$ are not the same map; they take different "paths". Let's trace out what happens: the top-right path is $a_{i,j}$, and the bottom-left path is $b_{i,j}$. $\require{AMScd}$
$$\require{AMScd}
\begin{CD}
\prod_{i\in I}F(U_i) @>{\Large p_i}>> F(U_i);\\
@V{\Large p_j}VV @VV{\Large\mathrm{res}_{U_i\cap U_j}^{U_i}}V \\
F(U_j) @>>{\Large\mathrm{res}_{U_i\cap U_j}^{U_j}}> F(U_i\cap U_j);
\end{CD}$$
The two maps $\alpha$ and $\beta$ to the product $\prod_{i,j\in I}F(U_i\cap U_j)$ agree on a specific element $(s_i)_{i\in I}$ of the product $\prod_{i\in I}F(U_i)$ if and only if $a_{i,j}$ and $b_{i,j}$ agree on it for all pairs of $i,j\in I$. But $a_{i,j}$ and $b_{i,j}$ are clearly going to be different maps in general, so this is not a trivial condition.
The constant presheaf fails to be a sheaf on a variety consisting of two points. For the constant presheaf the group of global sections is isomorphic to one copy of the constant group, whereas in the associated sheaf the group of global sections is isomorphic to the direct sum of two copies of the constant group.
This is essentially the same as your proof, which is correct.
Best Answer
I'm not sure exactly what $\cal F$ and $\cal G$ are, but let's takes some topological space $X$, let $\cal F$ be the sheaf of continuous complex functions on $X$ and $\cal G$ be the sheaf of continuous non-zero complex functions on $X$. Your definition then makes sense as a sheaf homomorphism.
This map exp is a surjective map of sheaves, but its image as a presheaf is not necessarily the whole of $\cal G$. If the presheaf image is $\cal H$, then ${\cal H}(U)$ is the set of functions on $U$ with a continuous logarithm. If say $X=\Bbb C$ then ${\cal H}(\Bbb C-\{0\})\ne{\cal G}(\Bbb C-\{0\})$ since the function $z\mapsto z$ is continuous and nonzero on $\Bbb C-\{0\}$ but has no continuous logarithm on that set.
But exp is surjective as a map of sheaves. The image of exp is the sheaf of functions on $X$ which locally have continuous logarithms, that is all nonzero continuous functions, since a continuous nonzero function has a continuous logarithm in some neighbourhood of any given point.