The explicit formula for cup product on group cohomology is as simple as can be. For simplicity let's consider integer coefficients $H^*(G;\mathbb{Z})$, although this works for any coefficients as long as they're untwisted.
Let's define group cohomology using inhomogeneous cochains; thus we take the abelian groups $C^n(G;\mathbb{Z}) :=$ functions from $G^n$ to $\mathbb{Z}$, endowed with a differential $d: C^n \to C^{n+1}$, and then $H^n(G;\mathbb{Z})$ is the usual cohomology $\ker d_n/\operatorname{im} d_{n-1}$.
Anyway, cup product is a map from $H^k(G) \otimes H^m(G)$ to $H^{k+m}(G)$, and it comes from a map $C^k(G) \otimes C^m(G)$ to $C^{k+m}(G)$. Namely, given two cochains $f: G^n \to \mathbb{Z}$ and $g: G^m \to \mathbb{Z}$, define
$$ f \wedge g: G^{k+m} \to \mathbb{Z} $$ by
$$ f\wedge g(x_1,...x_{k+m}) = f(x_1,...x_k)g(x_{k+1},...x_{k+m}) $$
You can check by hand that the differential interacts with this operation by
$$ d(f \wedge g) = df \wedge g + (-1)^k f \wedge dg $$
Thus this "wedge product" of cochains descends to a product on group cohomology, and this is exactly cup product. This is also how cup product is defined for de Rham cohomology; differential forms have a natural wedge productwedge product which satisfies $d(f \wedge g) = df \wedge g + (-1)^k f \wedge dg$, and so this induces the cup product on $H^*(M;R)$.
Topologically, cup product is the composition of
$$ H^k(Y) \otimes H^m(Y) \to H^{k+m}(Y \times Y) \to H^{k+m}(Y) $$
where the first map is the Kunneth map (just pullback by the two projections $Y \times Y \to Y$), and the second map is restriction to the diagonal. Applying this perspective to group cohomology, we would first define $f \times g : (G \times G)^{k+m} \to \mathbb{Z}$ by
$$ f \times g ((x_1,y_1),...(x_{k+m},y_{k+m})) = f(x_1,...x_k)g(y_{k+1},...,y_{k+m}). $$
Upon restriction to the diagonal $G < G \times G$, $f \times g$ restricts to $f \wedge g$ above.
