If $R > 0$ and $n \in \mathbb{Z}$ we can define the singular 1-cube $c_{R, n}\colon [0, 1] \rightarrow \mathbb{R}^2$ by
$$c_{R, n}(t) = (R\cos(2\pi n t), R\sin(2\pi n t))$$
We know that $c_{R, n} \ne \partial c$ for any 2-chain $c$ in $\mathbb{R}^2 - 0$. For assume $c_{R, n} = \partial c$ for some 2-chain $c$ in $\mathbb{R}^2$. Then
$$2\pi n = \int_{\partial c_{R, n}} d\theta = \int_{c_{R, n}}d(d\theta)) = 0$$
where $\theta$ is the 2nd coordinate function in the polar coordinate function (i.e. the one that gives us the angle of a point in $\mathbb{R}^2$) since $d\theta$ is closed.
I am confused, however, because if we define the 2-cube $c$ (in $\mathbb{R}^2 - 0$) by
$$c(s, t) = s\cdot c_{R, n}(t)$$
we get
$$\partial c(x) = -c(0,x) + c(1, x) + c(x, 0) - c(x, 1) = c_{R, n}(x)$$
since
$$c(x, 0) = x\cdot c_{R, n}(0) = x\cdot c_{R, n}(1) = c(x, 1)$$
I'm clearly doing something(s) wrong, and I'd like to understand what. Any help will be appreciated.
That really isn't much of an answer but ... as you asked for it ... well, the mistake is that the image of the function $$c:[0,1]\times[0,1]\to\mathbb{R}^2$$ given by $c(s,t)=s\cdot c_{R,n}(t)$ is not contained in $\mathbb{R}^2\setminus\{0\}$, because $c(0,t)=(0,0)$ for any $t\in[0,1]$. So, your construction of $c$ simply shows that $c_{R,n}$ is exact (i.e. it is the boundary of some $2$-chain) in $\mathbb{R}^2$, but doesn't allow you to conclude anything about what happens in $\mathbb{R}^2\setminus\{0\}$.
By the way, your definition of $c$ is essentially the construction of the cone on $c_{R,n}$ by means of the linear structure of $\mathbb{R}^2$, with vertex exactly in $(0,0)$.