It is a well-known fact in algebraic topology that for an abelian group $A$ (or more generally, any sheaf of abelian groups $A$) and any topological space $X$, the sheaf cohomology group $H^1 (X, A)$ is in bijection with the set of $A$-torsors (a.k.a. principal $A$-bundles) on $X$. Roughly speaking, a local trivialisation for an $A$-torsor gives a Čech $1$-cocycle for $A$, and the cohomology class of this cochain depends only on the isomorphism class of the $A$-torsor and not the choice of trivialisation. As I understand it, the definition of $H^1$ for non-abelian $A$ is designed so that this theorem remains true.
So how do we bring this idea to group cohomology? Well, for any discrete group $G$ and any $G$-module $A$, the group cohomology $H^* (G, A)$ is naturally isomorphic to topos cohomology $H^* (\mathbf{B} G, A)$, where $\mathbf{B} G$ is the category of all left $G$-sets. The definition of $A$-torsor makes sense in any topos and for any internal group $A$, and in the case of $\mathbf{B} G$ it comes down to this: an $A$-torsor is an inhabited left $G$-set $X$ equipped with a $G$-equivariant right $A$-action such that the map $X \times A \to X \times X$ sending $(x, a)$ to $(x, x \odot a)$ is a bijection.
In particular, for any choice of $x$, the map $a \to x \odot a$ is a bijection of sets $A \to X$. But this is not a $G$-equivariant bijection: after all, $g \cdot (x, a) = (g \cdot x, g \cdot a)$, and that gets mapped to $(g \cdot x, (g \cdot x) \odot (g \cdot a))$, which by $G$-equivariance is equal to $g \cdot (x, x \odot a) = (g \cdot x, g \cdot (x \odot a))$; so
$$(g \cdot x) \odot (g \cdot a) = g \cdot (x \odot a)$$
but it is not necessarily true that $x \odot (g \cdot a) \stackrel{?}{=} g \cdot (x \odot a)$. Regardless, we have a set-theoretic bijection, so for each $g$ in $G$ there exists a unique element $f(g)$ of $A$ such that
$$g \cdot x = x \odot f(g)$$
and given another $h$ in $G$, we have
\begin{align}
x \odot f (h g)
& = h g \cdot x \\
& = h \cdot (g \cdot x) \\
& = h \cdot (x \odot f(g)) \\
& = (h \cdot x) \odot (h \cdot f(g)) \\
& = (x \odot f(h)) \odot (h \cdot f(g)) \\
& = x \odot (f(h) (h \cdot f(g)))
\end{align}
which by uniqueness implies
$$f (h g) = f(h) (h \cdot f(g))$$
which lo and behold is the condition defining a $1$-cocycle for $A$. (You could do all this for a left $A$-action on $X$, but that gives a different equation.)
What if we choose a different element of $X$, say $x'$? Again, whatever $x'$ is, there is a unique element $c$ of $A$ such that
$$x' = x \odot c$$
and so if $f'$ is the $1$-cocycle associated with $x'$, then
$$g \cdot x' = x' \odot f'(g) = x \odot (c f'(g))$$
and
$$g \cdot x' = g \cdot (x \odot c) = (g \cdot x) \odot (g \cdot c) = x \odot (f(g) (g \cdot c))$$
and therefore
$$f'(g) = c^{-1} f(g) (g \cdot c)$$
which is precisely the definition of equivalence of $1$-cochains.
OK, you say, so every $A$-torsor gives rise to a cohomology class of $1$-cocycles, and obviously this construction depends only on the isomorphism class of the $A$-torsor. What about the other way around? Let $f : G \to A$ be a $1$-cocycle and let $X = A$ with the following $G$-action:
$$g \cdot x = x f(g)$$
This is a $G$-action, because $f(1_G) = 1_A$ and
$$h g \cdot x = x f(h g) = x f(h) (h \cdot f(g)) = (h \cdot x) (h \cdot f(g)) = h \cdot (x f(g)) = h \cdot (g \cdot x)$$
and obviously $A$ acts on itself $G$-equivariantly because $A$ is a $G$-group. Observe also that $X$ has a canonical element, namely $1_A$, and it is easy to see that the $1$-cocycle that it defines is exactly $f$. On the other hand, if $f$ is the $1$-cocycle corresponding to an $A$-torsor $\tilde{X}$ and an element $\tilde{x}$, then there is a unique $(G, A)$-equivariant isomorphism $X \to \tilde{X}$ sending $1_A$ to $\tilde{x}$. Thus the two constructions are mutually inverse up to isomorphism, and $H^1(G, A)$ really does classify $A$-torsors up to isomorphism.
