What I'm going to say is pretty much the same that JK34 has written in their answer, but in a more elementary approach that is hopefully adding some insight.
Suppose that you want to look at the "shape" of a group $G$. That is, let's construct a space that "looks like $G$". For simplicity suppose that $G$ is finite or finitely presented.
The idea is as follows: draw a point, call it $*$. Now for each element $g$ of $G$, draw a loop based at $*$, that is, a closed path from $*$ to itself.
The identity is assigned a "trivial" loop, meaning that it doesn't have any extension, it "stays at the base point" (imagine the trivial loop in algebraic topology).
Given $g\in G$, also $g^{-1}$ is in $G$. Don't add an extra path for $g^{-1}$, just imagine that your loops can be walked either way.
Now, given $g,h\in G$, clearly $gh$ is in $G$ too, so it has its own arrow. However, this arrow should not be independent of $g$ and $h$, it should be somehow the composition. A way to encode this (at least up to homotopy) is to add a 2-simplex (a filled triangle) between those three arrows (can't draw them here, sorry! Imagine a triangle with vertices all equal to $*$ and sides $g$, $h$, and $gh$). Do this for every pair of elements of $G$.
Now composition in the group is associative, that is, $(gh)i=g(hi)$ for each $g,h,i\in G$. Consider the triangles determined by $(g,h)$, $(gh,i)$, $(g,hi)$ and $(h,i)$. Can you see that these triangle form the faces of a tetrahedron? The first two and last two, pairwise glued, should again in some sense be the same, because of associativity. So, fill this tetrahedron too, with a 3-cell.
...and so on, imagine an $n$-dimensional simplex every time you have a composition of $n-1$ elements.
The resulting space is going to be a simplicial complex, called the classifying space of $G$, denoted by $BG$ (notation may vary).
Try to prove that
The classifying space of $\Bbb{Z}$ is the circle.
(Hint: we have one loop for each element of $G$...). If you understand that, it shouldn't come as a surprise that
More generally, the fundamental group of $BG$ is $G$
and (can you see why?),
All higher homotopy groups of $BG$ are trivial.
Now, as we have seen, a 2-simplex (i.e. triangle) in $BG$ is given by a pair $(g,h)$, with $g,h\in G$. What are the faces of that triangle? Those are the sides, which as we have constructed, are $g,h$, and $gh$. So, adding the signs consistently, it makes sense that the boundary of the simplex $(g,h)$ is
$$
\partial(g,h) = g - gh + h.
$$
This should give you an idea why the coboundary map in group cohomology is defined that way: we have that $(df)(g,h)=f(\partial(g,h))$.
Can you see what the boundary maps of simplices of degree $3$ will do? And higher?
This should also give you an idea why, in general,
The singular cohomology of $BG$ is exactly the group cohomology of $G$.
So, in short: group cohomology is helpful to study groups because it studies the "shape of a group" in the sense of algebraic topology.
The paper S. I. Adyan and V. S. Atabekyan, V. S.
Central extensions of free periodic groups,
Mat. Sb. 209 (2018), no. 12, 3–16; translation in
Sb. Math. 209 (2018), no. 12, 1677–1689
proves that if $n\geq 665$ is odd and $m\geq 2$, then the Schur multiplier $H^2(B(m,n),\mathbb Z)$ for the free Burnside group $B(m,n)$ of exponent $n$ on $m$-generators is free abelian of countable rank. In the arXiv version
this is Corollary 3 and so using the universal coefficients theorem, $H^2(B(m,n),\mathbb R^n)$ is quite huge. If you want an explicit class, then that is beyond my competency range.
Best Answer
Just stumbled across that old question while researching the subject on the internet, ended up doing the computation myself, and thought I would give an answer in case someone finds themself in the same situation.
A possible quasi-isomorphism $u_n: C^n(C_m,A)\to A$ from the standard complex to the periodic one is given by the following formulas: if $n=2r$, and $\tau$ is a generator of $C_m$, $$u_{2r}: f\mapsto (-1)^r\sum_{g_1,\dots,g_r\in C_m}\sum_{\varepsilon_1,\dots,\varepsilon_r\in \{0,1\}} (-1)^{\sum \varepsilon_i} f(g_1,\tau^{\varepsilon_1},\dots,g_r,\tau^{\varepsilon_r}),$$ and if $n=2r+1$, $$u_{2r+1}: f\mapsto (-1)^{r+1}\sum_{g_1,\dots,g_r\in C_m}\sum_{\varepsilon_1,\dots,\varepsilon_{r+1}\in \{0,1\}} (-1)^{\sum \varepsilon_i} f(\tau^{\varepsilon_1},g_1,\tau^{\varepsilon_2},\dots,g_r,\tau^{\varepsilon_{r+1}}).$$
In the example I was interested in, the action of $C_m$ on $A$ was trivial, but I read the computation carefully and I don't think it intervened in this formula.