What is the intuition between 1-cocycles (group cohomology)?

I believe there's a very simple motivation for the cocycle condition:

Suppose $f:X \to R$ is a function on a space $X$ with values in a ring $R$ (if you wish, $X$ is a manifold, and $R$ is $\mathbb{R}$ or $\mathbb{C}$, and the function $f$ is differentiable or even holomorphic). Suppose that $G$ is a group acting on $X$ (say on the left). Then for the function to be $G$-invariant is to say that $f(gx)=f(x)$ for all $x \in X$ and $g \in G$.

Often, this condition is too restrictive, i.e. there aren't enough functions satisfying this condition (as well as some other niceness conditions like holomorphic or whatever). Instead, you might want to condition the class of functions satisfying the following relation $$f(gx)=j(g,x) f(x),$$ where $j:G \times X \to R$ is a function. I'll give motivation for this type of relation below, but first I'd like to mention the consequences of such a relation.

Suppose there exists a nonzero $f$ satisfying this relation. Then for all $x \in X$ and $g_1,g_2 \in G$, we have $$f(g_1 g_2 x) = j(g_1 g_2, x) f(x).$$ On the other hand, we have

\begin{eqnarray*} f(g_1 g_2 x) &=& j(g_1,g_2 x) f(g_2 x) &=& j(g_1,g_2 x ) j(g_2,x) f(x) \end{eqnarray*}

Thus we find that $j(g_1 g_2, x) = j(g_1,g_2 x) j(g_2, x)$. If we let $\mathcal{O}^\times$ denote the group of nonzero functions on $X$ under multiplication (possibly only those satisfying some condition, like continuity or differentiability or holomorphicity), then we can think of $j$ as a map $G \to \mathcal{O}^\times$. We can think of $G$ as acting on $\mathcal{O}^\times$ on the right by precomposition, and then this condition on $j$ is precisely the cocycle condition for a map $G \to \mathcal{O}^\times$. In particular, it determines an element of $H^1(G,\mathcal{O}^\times)$.

So whence the condition $f(gx)=j(g,x) f(x)$? I'll start with some classical though possibly less conceptual motivation from special functions, then give the geometric interpretation in terms of line bundles.

Let $L$ be a lattice in the complex plane, i.e. the $\mathbb{Z}$-span of a $\mathbb{R}$-basis of $\mathbb{C}$. A theta function is a meromorphic function such that $\theta(z+\omega)=j(\omega,z)\theta(z)$ for all $z \in \mathbb{C}$, $\omega \in L$. More generally, if $X$ is a contractible Riemann surface, and $G$ is a group which acts on $X$ under sufficiently nice conditions, consider meromorphic functions $f$ on $X$ such that $f(gz)=j(g,z)f(z)$ for $z \in X$, $g \in G$, where $j: G \times X \to \mathbb{C}$ is holomorphic for fixed $g$. In the case of theta functions, $G$ is $L$, and $X$ is $\mathbb{C}$.

Another basic example is a modular form such as $G_{2k}(z)$, which satisfies $G_{2k}(g z) = (cz+d)^{2k} G_{2k}(z)$, where $g= \left(\begin{array}{cc} a & b \\ c & d \end{array}\right) \in G = SL_2(\mathbb{Z})$ acts as a fractional linear transformation. It follows automatically that something as simple as $(cz+d)^{2k}$ is a cocycle in group cohomology, since $G_{2k}$ is, for example, nonzero. In this case $X = \mathcal{H}$, the complex upper-half plane.

Now for some geometric motivation. I'll stick to the case of elliptic curves, though I might rewrite this and make it more general. We define an elliptic curve to be $E=\mathbb{C}/L$ for a two-dimensional lattice $L$. Note that the first homology group of this elliptic curve is isomorphic to $L$ precisely because it is a quotient of the universal cover $\mathbb{C}$ by $L$. We will see that a theta function is a section of a line bundle on an elliptic curve. Since any line bundle can be lifted to $\mathbb{C}$, the universal cover, and any line bundle over a contractible space is trivial, the line bundle is a quotient of the trivial line bundle over $\mathbb{C}$. We can define a function $j(\omega,z):L \times \mathbb{C} \to \mathbb{C} \setminus \{0\}$. Then we identify $(z,w) \in \mathbb{C}^2$ (i.e. the line bundle over $\mathbb{C}$) with $(z+\omega,j(\omega,z)w)$. For this equivalence relation to give a well-defined bundle over $\mathbb{C}/L$, we need the following: Suppose $\omega_1,\omega_2 \in L$. Then $(z,w)$ is identified with $(z+\omega_1+\omega_2,j(\omega_1+\omega_2,z)w$. But $(z,w)$ is identified with $(z+\omega_1,j(\omega_1,z)w)$, which is identified with $(z+\omega_1+\omega_2,j(\omega_2,z+\omega_1)j(\omega_1,z)w)$. In other words, this forces $j(\omega_1+\omega_2,z) = j(\omega_2,z+\omega_1)j(\omega_1,z)$. This means that, if we view $j$ as a function from $L$ to the set of non-vanishing holomorphic functions $\mathbb{C} \to \mathbb{C}$, with (right) L-action on this set defined by $(\omega f)(z) \mapsto f(z+\omega)$, then $j$ is in fact a $1$-cocyle in the language of group cohomology. Thus $H^1(L,\mathcal{O}(\mathbb{C}))$, where $\mathcal{O}(\mathbb{C})$ denotes the (additive) $L$-module of holomorphic functions on $\mathbb{C}$, classifies line bundles over $\mathbb{C}/L$. What's more is that this set is also classified by the sheaf cohomology $H^1(E,\mathcal{O}(E)^{\times})$ (where $\mathcal{O}(E)$ is the sheaf of holomorphic functions on $E$, and the $\times$ indicates the group of units of the ring of holomorphic functions). That is, we can compute the sheaf cohomology of a space by considering the group cohomology of the action of the homology group on the universal cover! In addition, the $0$th group cohomology (this time of the meromorphic functions, not just the holomorphic ones) is the invariant elements under $L$, i.e. the elliptic functions, and similarly the $0$th sheaf cohomology is the global sections, again the elliptic functions.

Once one constructs a line bundle on $E$ as a quotient of a line bundle on $\mathbb{C}$ via the cocycle $j$, a section of that line bundle corresponds to a function $f$ on the complex plane satisfying the cocycle condition. Similarly, modular forms are sections of line bundles on modular curves.

Here's on simple interpretation of $H^1(G, A)$ that you can check directly.

If $G$ acts on $A$, then we may construct a semidirect product $G \ltimes A$. Then $H^1(G,A)$ corresponds to the conjugacy classes of complements to $A$ in the semidirect product. (In general, $H^1(G,A)$ only a pointed set - a set with a distinguished element (the identically 1 function); it is not a group unless $A$ is abelian.)

Proof: Any complement must be of the form $\{(g, a(g)) : g \in G\}$ for some $a : G \to A$. Check that the condition for such a function $a$ to be a complement is precisely the cocycle condition, and that if you change a cocycle by a coboundary, that corresponds to conjugation. The "special" cocycle that is identically 1 corresponds to the original complement $G$.