Let $U_1$ be an open subset of $\Bbb R^n$ and $U_2$ be an open subset of $\Bbb R^m.$ Let $f : U_1 \longrightarrow U_2$ be a smooth function. Then for any $p \geq 1$ we can define a function $f^* : \Omega^p (U_2) \longrightarrow \Omega^p (U_1)$ by $$f^* (\omega) (x) (\xi_1, \xi_2, \cdots, \xi_p) : = \omega (f(x)) \left (Df(x)(\xi_1), Df(x)(\xi_2), \cdots , Df(x)(\xi_p) \right ).$$
Now let $d \theta = \frac {-y} {x^2 + y^2}\ dx + \frac {x} {x^2 + y^2}\ dy$ be an $1$-form in $\Bbb R^2 \setminus \{0\}.$ Let us define the $1$-form $c_{R,n} : [0,1] \longrightarrow \Bbb R^2 \setminus \{0\}$ by $$c_{R,n} (t) = \left (R \cos (2\pi n t), R \sin (2 \pi n t) \right ),\ t \in [0,1].$$
Then $c_{R,n}^* : \Omega^1 (\Bbb R^2 \setminus \{0\}) \longrightarrow \Omega^1([0,1])$ will be defined as $$c_{R,n}^* (\omega) (t) (\xi) = \omega (c_{R,n} (t)) (Dc_{R,n} (t) (\xi)).$$ With this definition in mind if we calculate $c_{R,n}^* (d\theta) (t) (\xi)$ then what do we get?
We should get $$\begin{align*} c_{R,n}^* (d\theta) (t) (\xi) & = d\theta(c_{R,n} (t)) (Dc_{R,n} (t) (\xi)) \\ & = d\theta (R \cos (2 \pi n t), R \sin (2 \pi n t)) (2 \pi n \xi (-\sin (2 \pi n t), \cos (2 \pi n t))) \\ & = 2 \pi n\ dt\ ((2 \pi n \xi (-\sin (2 \pi n t), \cos (2 \pi n t))) \end{align*}$$
Now how do I compute $dt\ ((2 \pi n \xi (-\sin (2 \pi n t), \cos (2 \pi n t)))\ $? Any help in this regard will be warmly appreciated.
Thanks for your time.
EDIT $:$ For a $(k-1)$-form $\omega$ we define $k$-form $d\omega$ in the following way $:$ $$d \omega (x) (\xi_1,\xi_2, \cdots, \xi_{k+1}) = \sum\limits_{l=1}^{k+1} (-1)^{l-1} D\omega(x) (\xi_1,\xi_2, \cdots, \hat {\xi_l}, \cdots, \xi_{k+1}),$$ where $\hat {\xi_l}$ means that the $l$-th coordinate is missing.
With that definition we have $$dt\ ((2 \pi n \xi (-\sin (2 \pi n t), \cos (2 \pi n t))) = 2 \pi n \xi\ (\cos (2 \pi n t) + \sin (2 \pi n t)).$$ But then what is $c_{R,n}^* (d \theta)\ $? By definition $c_{R,n}^*$ should be a $1$-form defined on $[0,1].$