Examples of surjective sheaf morphisms which are not surjective on sections

Take $X=\mathbb R$ for your topological space, the constant sheaf $\underline {\mathbb Z} $ for $\mathcal F$ and for $\mathcal G$ the direct sum of two skyscraper sheaves with fibers $\mathbb Z$ at two distinct points $P,Q\in \mathbb R$, that is $\mathcal G=\mathbb Z^P \oplus \mathbb Z^Q$.
The natural restriction $\mathcal F=\underline {\mathbb Z} \to \mathcal G=\mathbb Z^P \oplus \mathbb Z^Q$ is a surjective sheaf morphism but the associated group morphism on global sections $\mathcal F(X)=\mathbb Z \to \mathcal G (X)=\mathbb Z \oplus \mathbb Z$ is not surjective [its image is the diagonal of $\mathbb Z \oplus \mathbb Z$, consisting of pairs $(z,w)$ with $z=w$].

Edit This example can easily be adapted to a three point space space: thanks to Pierre-Yves who, in a comment to Alex's answer, suggested that.

Take $X=\{P,Q, \eta\}$ with closed sets $X,\emptyset, \{P\}, \{Q\}, \{P,Q\}$ (this is the same space as Alex's).The rest is exactly the same as above. Namely $\mathcal F=\underline {\mathbb Z} $, $\mathcal G=\mathbb Z^P \oplus \mathbb Z^Q$, $\mathcal F=\underline {\mathbb Z} \to \mathcal G=\mathbb Z^P \oplus \mathbb Z^Q$ the restriction, which is again a surjective sheaf morphism, and $\mathcal F(X)=\mathbb Z \to \mathcal G (X)=\mathbb Z \oplus \mathbb Z \:$ not surjective (the image being again the diagonal of $\mathbb Z \oplus \mathbb Z$).
The main point is that the stalks of $\mathbb Z^P$ are:
$(\mathbb Z^P)_P=\mathbb Z,(\mathbb Z^P)_Q=0, (\mathbb Z^P)\eta=0$, because $P$ is a closed point. Ditto for $\mathbb Z^Q$.

Tangential remark It might be of some interest to notice that there is a scheme structure on $X$ which makes it the smallest possible non affine scheme. This is explained in the book The Geometry of Schemes by Eisenbud and Harris, on page 22.

Let $M$ be a smooth manifold, and consider the sheaf of closed 1-forms. There is a surjection from the sheaf of smooth functions to the sheaf of closed 1-forms (namely, the exterior derivative, $f \mapsto df$), which is surjective in the category of sheaves (by the Poincare lemma), but which is in general not surjective (if $M$ has nontrivial first de Rham cohomology).

Let me try as elementary as is humanly possible:

$X=\{p,q_1,q_2\}$, consisting of only three elements! The open sets are $U_0=\{p\}$, $U_1=\{p,q_1\}$, $U_2=\{p,q_2\}$, and of course the empty set and $X$. Define $\mathscr{F}(U)=\mathscr{G}(U)=\mathbb{Z}$ on all non-empty open sets $U$. Now, the trick is going to be in the restriction maps and the morphism. Define all the restriction maps on $\mathscr{G}$ to be the identity, but the restrictions $\mathscr{F}(X)\rightarrow \mathscr{F}(U_i)$, $i=1,2$ are multiplication by 2. The restrictions $\mathscr{F}(U_i)\rightarrow \mathscr{F}(U_0)$ are again the identity maps (this forces the restriction from $X$ to $U_0$ to also be multiplication by 2).

Define $\phi:\mathscr{F}\rightarrow \mathscr{G}$ to be the identity on all open sets except for $X$, where it is multiplication by 2. Then $\phi(X)$ is not surjective, but it is surjective on all the stalks (check!). I hope I haven't made a mistake.