Why is a section of a sheaf over closed set defined this way? Why is a section of a sheaf $F$ over closed set $S \subset X$ is defined as inductive limit $$ \varinjlim_{S\subset U} F(U)\; ?$$
From my point of view, we should define it as a function, which each point $x \in S$ maps to $F_x$, such that $S$ can be covered by open sets $U_i$, and there are sections $s_i \in F(U_i)$, which coincide in stalks. Is it equivalent? Maybe, for sufficiently good topological spaces?
 A: Your point of view  for the definition of $\Gamma (S, F)$ and thus implicitly for $F\mid S $ is indeed the correct one.         
The formula   $\Gamma (S, F)= \varinjlim_{S\subset U} F(U)$ should not be taken as the definition, even though it is is  true in some cases, for exmple   if $X$ is paracompact: cf. Corollaire 1 to Théorème 3.3.1, page 151 of Godement's Topologie Algébrique et Théorie des Faisceaux.
A: (I am very new to this, so please don't accept anything written here without thinking about it yourself! These are just my thoughts on this identity and how I might show it.) You can think of a sheaf as a bundle of germs $\Lambda F$ sitting over $X$, with a projection map $p:\Lambda F \rightarrow X$ sending $p(germ_x f) = x$. This bundle is topologized so that  all the sections on $U$ of the form $s_f(x) = germ_x f$ for $f \in F(U)$ are continuous. (The images of these maps are actually taken as a subbasis for the topology, if I recall.) Sections of $F$ over $S$ then correspond to continuous lifts of the inclusion $i:S \hookrightarrow X$ through $p$. I believe each lift corresponds exactly to one of the functions you describe, so we can simply think of lifts. 
The question now becomes: can we extend lifts to an open set about $S$? Perhaps this is where the paracompactness condition enters. Given a lift $l$ from $S$ up into the bundle, we obtain lots of information about it in the form of open sets about points in $S$ on which $l$ looks like a map of the form $l(x) = germ_x f$. Perhaps we can then control this using paracompactness and extend $l$ to an open set about $S$, giving the desired equality.
