Can every element in the stalk be represented by a section in the top space? Let $S$ be a sheaf over $X$ and $r$ an element in $S_x$ for some $x$ in $X$. Must there exist a section $s$ in $S(X)$ such that such that $s$ equals $r$ when mapped to $S_x$ by the canonical map?  
 A: This is only true in pathological cases. Neither in complex analysis (see above) nor in algebraic geometry: If we take the structure sheaf $\mathcal{O}_{Spec(A)}$ of an affine scheme, then the question is whether the localzation maps $A \to A_{\mathfrak{p}}$ are surjective, which is almost never true.
A: No: take the sheaf of holomorphic functions $S=\mathcal O$ on $X=\mathbb C$. Then the stalk at $x=1$ of $1/z$ does not come from $\mathcal O(\mathbb C)$.
@Arturo: no you are not misremembering. Here is an example. Take $X=\mathbb P_n (\mathbb C)$ and $S=\mathcal O (-1)$= tautological bundle. Then every stalk $S_x$ is isomorphic with $\mathcal O_x$ so is not zero but $S(\mathbb P_n (\mathbb C))=0$. (By the GAGA principle of Serre you can choose to think in holomorphic way or algebraic way in this example)
@Robin Chapman: Yes, there are sheaves $S$ with all morphisms $S(X) \to S_x$ surjective, that are not flabby. For example, take $X=\mathbb R$ and $S=C$ = sheaf of continuous functions. Then  for $x\in \mathbb R$ and $f_x \in C_x$ take a bump function $\phi$ equal to $1$ in vicinity of $x$ and support smaller than the domain of definition of $f$. Then $\phi f$ (extended by zero) is a global function in $C(\mathbb R)$ and its germ at $x$ is equal to $f_x$. But of course $C$ is not flabby.
A: This raises another question. If the restriction maps from $S(X)$ to $S(U)$
are surjective for all open sets $U$ (a sheaf with this
property is called flasque or flabby) then certainly the maps
from $S(X)$ to each stalk $S_x$ are surjective. So are there sheaves
$S$ with the property that each map $S(X)\to S_x$ is surjective but which are
not flabby? I expect there are, but cannot see an example immediately.
