Let $c_{R,n}: [0,1]\to \mathbf R^2$ be the map $t\mapsto (R\cos 2\pi n t,R\sin 2 \pi n t)$. I'm trying to compute the pullback of the 1-form $\omega=-\frac{y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$ on $\mathbf R^2$ under this map. (This is part of an exercise from Spivak, so I use his technique and notation.)
Write $c=c_{R,n}$ for brevity. Then $(c^\ast\omega)(p)=c^\ast(w(c(p))$. Explicitly, $$(c^\ast\omega)(p)(\tau)=c^\ast(\omega(c(p)))(\tau)=w(c(p))(c_\ast \tau)$$ where $c_{\ast}$ is the differential of $c$ (I guess at some $t$ -- Spivak's notation is strange since $c_\ast$ must depend on the point).
I have $c_\ast\tau=(-2\pi n R \tau \sin 2 \pi n t, 2\pi n R \tau \cos 2 \pi n t)^T$, and the problem boils down to apply $$-\frac{\sin 2 \pi n t}{R}dx+\frac{\cos 2 \pi n t}{R}dy$$ to the above vector (if my computations are correct). But I don't know how to do that -- Spivak doesn't seem to talk about this.
So, how do I do that? (The answer should be $2\pi n d t(\tau)$.)
Call $c_1$ and $c_2$ the two coordinates of your map.
Then, in the expression of $\omega$:
Step 1. Replace $x$ and $y$ by the expressions of $c_1$ and $c_2$ respectively.
Step 2. Replace d$x$ and d$y$ by $\partial c_1\over \partial t$ and $\partial c_2\over \partial t$ respectively.
Step 3. Add d$t$ in the end.
What you are missing in your solution is that in the end you should think of $$-\frac{\sin 2 \pi n t}{R}dx+\frac{\cos 2 \pi n t}{R}dy$$ as a $1\times 2$ matrix that represents your linear map, so that applying the linear map consists of a dot product.