Let $f: X \rightarrow Y$ be a continuous map of topological spaces, and $F$ a sheaf of rings on $X$. The direct image sheaf $f_{\ast}F$ on $Y$ is given by the formula $V \mapsto F(f^{-1}V)$. If $x \in X$, is it true in general that $F_x \cong (f_{\ast}F)_{f(x)}$?
We have $$(f_{\ast}F)_{f(x)} = \varinjlim\limits_{V \ni f(x)} F(f^{-1}V) = \varinjlim\limits_{f^{-1}V \ni x} F(f^{-1}V)$$ so it appears that this limit equals $F_x = \varinjlim\limits_{U \ni x} F(U)$ if for any neighborhood $U$ of $x$, there exists a neighborhood $V$ of $f(x)$ such that $f^{-1}V \subseteq U$.
Obviously this last statement isn't true for all continuous maps. For example I could take $X = \{a,b\}$ in the discrete topology, $Y = \{a,b\}$ in the indiscrete topology, and $f: X \rightarrow Y$ the map $a,b \mapsto a$. Then I can take $U = \{a\}$ (and $x = a$).
It appears that there is at least a canonical homomorphism $(f_{\ast}F)_{f(x)} \rightarrow F_x$.
This isn't true in general, for exactly the reason you mention. For instance, suppose $Y$ has only one point. Then $(f_*F)_{f(x)}=F(X)$, and you can easily find a sheaf $F$ on some space $X$ and a point $x$ such that $F(X)$ is not isomorphic to $F_x$.