Let $X$ be the two element set $\{0, 1\}$, and make $X$ into a topological space by taking each of the four subsets to be open. A sheaf $\mathscr{F}$ on $X$ is thus a collection of four sets with certain maps between them. Find the stalks $\mathscr{F}_0$ and $\mathscr{F}_0$.
To find $\mathscr{F}_0 = \varinjlim_{0 \in U} \mathscr{F}(U)$ we see that only $\{0\}$ and $X$ contain $0$.
So we have and indexing category $I$ with two objects $X$ and $\{0\}$ and morphism $$\{0\} \hookrightarrow X.$$ Now $\mathscr{F}$ is a contravariant functor so we get $$\mathscr{F}(X) \to \mathscr{F}(\{ 0 \})$$ in the category of whatever $\mathscr{F}(U)$'s belong to. Now I think that we want to make the diagram $$\require{AMScd} \begin{CD} \mathscr{F}(X) @>\alpha>> \mathscr{F}(\{0\})\\ @. {_{\rlap{\ \beta\circ \alpha}}\style{display: inline-block; transform: rotate(30deg)}{{\xrightarrow[\rule{4em}{0em}]{}}}} @VV\beta V\\ @. \mathscr{F}_0 \end{CD}$$ commute and also satisfy the universal property that the direct limit requires. Is there a way to explicitly compute $\mathscr{F}_0$ using this?
We can take $\mathscr{F}_0=\mathscr{F}(\{0\})$ and $\beta=1_{\mathscr{F}(\{0\})}$.
Then a cocone of the diagram consists of a pair of maps:\begin{eqnarray*} u\colon \mathscr{F}(\{0\})&\to& A\\ v\colon \mathscr{F}(X) &\to& A \end{eqnarray*} with $v=u\alpha $.
Our induced map $\mathscr{F}_0\to A$ is then just $u$. As $\beta=1_{\mathscr{F}(\{0\})}$, we have:
\begin{eqnarray*}u\beta\alpha=u\alpha=v\\ \\ {\rm and} \\ \\u\beta=u.\end{eqnarray*}
Conversely $u$ is the only map $w\colon \mathscr{F}(\{0\})\to A$ such that $w\beta=u$. Thus the universal property is satisfied.