Recover the stalk at a certain point along the base. Let $X$ be a topological space; its open sets form a directed set by the rule
 $$U\prec V\quad\textrm{iff}\quad V\subseteq U$$
If $\mathscr F$ is a sheaf, for example of rings, then by definition, for every $x\in X$ we have that
$$\mathscr F_x=\varinjlim_{U\ni x}\mathscr F(U)$$
where clearly $U$ is varying along all open neighborhoods of $x$.

Now if $\mathcal B$ is a base for the topology, intuitively one should have that 
$$\mathscr F_x=\varinjlim_{\substack {B\in\mathcal B\\B\ni x}}\mathscr F(B)$$
because $\{\mathscr F(B)\}$ looks like as a sort of subsequence of $\{\mathscr F(U)\}$. However I cannot make the above intuitive statement in a formal proof. 
 A: The key is that "only the really small open sets matter". Let then $F_x$ be the usual stalk at $x$ with canonical inclusions $i_U:F(U)\to F_x$, whereas $F_{(x)}$ is the "basic" stalk at $x$ has inclusions $j_B:F(B)\to F_{(x)}$.
On the one hand basic open sets are open sets and therefore we have for each $B\in \mathcal B$ a map $i_B:F(B)\to F_x$, so the universal property of the colimit gives us a map $F_{(x)}\to F_{x}$.
On the other hand, since $\mathcal B$ is a basis for the topology, for any $U$, there exists some $B\subseteq U$ a basic open set.
Now consider $\mathcal B_U$ the family of basic open sets contained in $U$, then for each $B\in \mathcal B_U$ we have a map $\mu_B:F(U) \to F(B)$ that naturally commutes with restrictions (that is $\mu_C = r_{B,C}\circ\mu_B$). This gives a map $F(U)\to F_{(x)}$ defined by $j_B\circ\mu_B$ which does not depend on the choice of $B$, this family gives rise to a map $F_x\to F_{(x)}$ (by the universal property pf the colimit).
Lastly I couldn't piece out how to do this in general but we are left to show that the maps $\alpha:F_x\rightleftarrows F_{(x)}:\beta$ are inverse to each other, the setting I chose is when $F$ is a sheaf of sets.
In this setting we have a representation of the stalks as the set of germs/"basic" germs at $x$
$$F_x = \coprod_{U\ni x} F(U)\big/\sim \qquad 
F_{(x)} = \coprod_{B\ni x} F(B)\big/\sim $$
where $s_U \sim s_V$ iff there exists $W\subseteq U\cap V$ such that $r_{U,W}(s_U) = r_{V,W}(s_V)$ (in both cases where $U,V,W$ are general open sets or just basic open sets).
Thus take an element $s\in F_x$ it is a germ of some element $s_U\in F(U)$ therefore it restricts to an element $s_B\in F(B)$ for some basic open set $B$ containing $x$, thus determining a "basic germ", since we can take $U$ to be a basic open set itself, this map $F_x\to F_{(x)}$ is surjective. Now take two germs $s,t\in F_x$ so that $\alpha(s) = \alpha (t)$, furthermore let $s_U$ and $t_V$ be the representatives of $s$ and $t$ respectively, then $\alpha(s)$ and $\alpha(t)$ have representatives $s_{B}$ and $t_{B'}$ in $F_{(x)}$, and since they are the same, itstands to rason to conclude there exists some basic open set $D\subseteq B\cap B'$ so that $r_{B,D}s_B = r_{B',D}t_{B'}$ but $s_B = r_{U,B}s_U$ and $t_{B'}=r_{V,B'}t_V$ thus $r_{U,D}s_U= r_{V,D}t_V$ and so $t=s$. Therefore $\alpha$ is injective and surjective which is an isomorphism in $\mathbf{Set}$.
