Explicit description of the elements of a local system If I have a topological space $X$ with a universal cover $p:\tilde{X}\to X$ and the fundamental group $G=\pi_1(X)$, and (say) $A$ is some $G$-module and $A_{\tilde {X}}$ is the constant sheaf defined on this universal cover for $A$, then (as far as I can tell) this is supposed to be enough information to define a local system $\mathcal{A}$, which is a sheaf on on $X$ itself. But I can't make sense of any of the explicit definitions of that I can find for what that sheaf actually is. By which I mean, exactly what the elements of the image of some given open set under the sheaf are in terms of the constant sheaf on the universal cover.
As an example, and with some notation translating, Wikipedia's description of a local system seems to say that for any open subset $U$ of $X$, $\mathcal{A}(U)=\lbrace a\in A_{\tilde{X}}(p^{-1}(U)):g \circ a=g\cdot a \rbrace \text{ for every }g\in G$, but what does it mean to "compose" an element of $g$ with an element of $a$ here? I understand that $g$ can be thought of as just the deck transformation map $\tilde{X}\to \tilde{X}$, but $a$ is supposed to be an element of $A$, not any kind of map on $\tilde{X}$, right? Or, digging deeper, the reference that page seems to be getting that from is this paper, which appears to say around page 7 (again with some translation of the notation) that $\mathcal{A}(U)=\lbrace a \in A_{\tilde{X}}(p^{-1}(U)):\text{ for every } x\in p^{-1}(U) \text{ and }g\in G, a(g \cdot x) = g \cdot a(x) \rbrace$. This seems to once again assume that elements of $A_\tilde{X}$ are some kind of functions from $\tilde{X}$ to somewhere else, but that doesn't make any sense to me at all.
My (admittedly limited) understanding of (pre)sheaves is that they're supposed to be functors from the collection of all the open sets of the space to some other category, in this case to the category of $G$-modules, so how can some element of the image of a functor like that also be a map from an element of the open set the functor came from to a topological space?
 A: $\mathcal{A}$ is a sheaf of abelian groups (or vector spaces), not a sheaf of $G$-modules, and an easy way to answer your questions is that : yes sections of a sheaf can be seen as some kind of functions. Let me expand a bit on this and then take a look at the definitions you linked.
If you have a very general sheaf $\mathcal{F}$ on a space $X$, a section $s$ indeed define a "function" $x\mapsto s_x$. The source is $X$, but the target is not a single set, instead, for any $x\in X, s_x\in\mathcal{F}_x$. Yet these "functions" very much behave as usual functions, and in particular if two "functions" agree then their corresponding section are the same. I am just saying here that if for any $x\in X, s_x=t_x$ then $s=t$.
But here, you have much more : $A_{\tilde{X}}$ is a constant sheaf and the best way to define a constant sheaf is indeed as a sheaf of usual functions : $A_{\tilde{X}}(U)$ is the set of locally constant function $U\to A$. So if $s\in A_{\tilde{X}}(U)$ is any section $s(x)$ simply means the value of $s$ at $x$ and will be an element of $A$.
Note that the two descriptions are not incompatible since $A_{\tilde{X},x}$ is canonically isomorphic to $A$ and $s_x=s(x)$ using this identification.

So now, let us have a look at the two definitions of the local sheaf associated to a $\pi_1(X,x)$-module $A$. Well in fact this is exactly the same definition (but the wikipedia article contains an error : it should be $a\circ g$ instead of $g\circ a$).
As you said, any $g\in G$ can be seen as deck transformation $g:\tilde{X}\to\tilde{X}$, and for any $U\subset X$, it restricts to a transformation $g:p^{-1}(U)\to p^{-1}(U)$. If $a$ is a locally constant function on $p^{-1}(U)$ then $a\circ g$ is also locally constant and thus define a new section $a\circ g\in\mathcal{A}(U)$. You can also use the $G$-structure on $A$ : if $a$ is a locally constant fonction on $p^{-1}(U)$, then for any $x\in p^{-1}(U), a(x)$ is a element of $A$ so it makes sense to write $g.a(x)$. The correspondance $x\mapsto g.a(x)$ is a locally constant function written $g.a$. So finally : $$\begin{align}\mathcal{A}(U)&=\{a: p^{-1}(U)\to A \text{ locally constant}~|~ a\circ g=g.a\}\\
&= \{a: p^{-1}(U)\to A \text{ locally constant}~|~ \forall x\in p^{-1}(U),  a(g.x)=g.a(x)\}\end{align}$$

An example : let $X=S^1=\mathbb{R/Z}$ bet the circle and $p:\mathbb{R}\to S^1$ its universal cover. We have $G=\mathbb{Z}$ and let $A=\mathbb{Z}$. We define a $G$-module structure on $A$ by putting $g.a=(-1)^ga$ (recall that $a,g\in\mathbb{Z}$) (Actually, we don't have much choice for the module structure here, this is the only non trivial one).
Now let $\mathcal{A}$ be the local system on $S^1$ associated to this module structure. We basically have two kinds of open set for which we want to compute $\mathcal{A}(U)$ :

*

*$U=S^1$ is the whole space. Then $p^{-1}(S^1)=\mathbb{R}$. A locally constant function $\mathbb{R}\to A$ is constant and completely determined by its image still denoted $a\in A=\mathbb{Z}$. The composition $a\circ g$ will always yield the same constant function $a$. However $g.a=(-1)^ga$ is not equal to $a$ for odd $g$ and non-zero $a$. It follows that the only $a$ such that for any $g, a\circ g=g.a$ is zero : $\mathcal{A}(X)=\{0\}$.

*$U=(x,y)$ is a small interval, then $p^{-1}((x,y))=\coprod_{n\in\mathbb{Z}}(x+n,y+n)$ with no overlapping. Then a locally constant function $a:p^{-1}((x,y))\to A$ will be constant on any $(x+n,y+n)$ and thus corresponds to the data for all $n\in\mathbb{Z}$ of $a_n\in A$ the constant value of $a$ on $(x+n,y+n)$. Now $a\circ g=g.a$ here means that $a_{n+g}=(-1)^ga_n$ (since the deck transformation $g$ corresponds to the translation $+g:\mathbb{R\to R}$). It follows that $a_g=(-1)^ga_0$ and thus $a_n$ is completely determined by $a_0$. Hence $\mathcal{A}((x,y))=A$.

Hence we have $\mathcal{A}((0.1, 0.9))=\mathbb{Z}$ and $\mathcal{A}((0.8, 1.2))=\mathbb{Z}$ and the intervals $(0.1,0.9), (0.8, 1.2)$ cover $S^1$. Consider the two sections corresponding to 1 under the above identifications. Can you see that they do not glue ?
