Calculating the pullback of a differential form in $\mathbb{R}^n$ In my course we defined the pullback $f^{*}\omega$ of a differential $k$-Form $\omega: V \subseteq \mathbb{R}^n \to \bigwedge^k\left(\mathbb{R}^n\right)$ via a differentiable map $f: U \subseteq \mathbb{R}^m \to V$ as 
$$
(f^{*}\omega)(x)[v_1, ..., v_k] := \omega(f(x))[f'(x)v_1, ..., f'(x)v_k].
$$
However, I have some trouble understanding this definition. In particular I don't quite understand what exactly is meant by the terms $f'(x)v_i$. I suppose these vectors are supposed to be the new inputs for the resulting $k$-Form over $U$. Let me showcase my confusion with the help of an example. Let $\omega = xdx + ydy + zdz$ and $f:[0,2\pi]\to\mathbb{R}^3$ is the curve $f(t) = (e^{t\sin{t}},t^2 - 2\pi t, \cos{\frac{t}{2}})$.
Then we have $f'(t) = ((\sin{t} + t\cos{t})e^{t\sin{t}}, 2t - 2\pi, -\frac{1}{2}\sin{\frac{t}{2}})$ and the above definition should reduce to:
$$
(f^{*}\omega)(t)[v] = \omega(f(t))[f'(x)v]
$$
where $v = (x,y,z) \in \mathbb{R}^3$. Going further:
$$
(f^{*}\omega)(t)[v] = \left(e^{t\sin{t}}dx + (t^2 - 2\pi t)dy + \cos{\frac{t}{2}}dz\right)[f'(x)v].
$$
Now, to me the only sensible way to calculate $f'(t)v$ is calculating the dot product $\langle f'(t), v\rangle$, since both $f'(t)$ and $v$ are vectors of $\mathbb{R}^3$ and don't really know of another way to multiply two of those (i think the cross product is out of question here). However, the dot product yields only a scalar, but the $1$-Form is waiting for another 3-vector. So this can't be right. What is the correct way to calculate this pullback?
 A: You should note that $f'$ (in your notation) is the total derivative of the differentiable map $f:U\to V$. Typically, this is written $df$ or $Df$. Indeed, at a point $p\in U$, $df_p:T_pU\to T_pV$ transforms tangent vectors to tangent vectors. In local coordinates, $df_p$ is the Jacobian matrix of the transformation - i.e. the matrix of first partials. 
It's good that you tried an example, and you are somewhat close. The only thing you are missing is how the total differential $df$ acts. Indeed, it acts as a linear transformation on elements of the tangent space. So, let's take $v\in T_t\mathbb{R}$ in your example. Because $v$ is in the tangent space to $\mathbb{R}$, this is of course just a scalar, i.e. a real number. Then, as you correctly calculated: 
$$ df_t = ((\sin{t} + t\cos{t})e^{t\sin{t}}, 2t - 2\pi, -\frac{1}{2}\sin{\frac{t}{2}}).$$
We apply this to $v$ to get $df_t(v)= ((\sin{t} + t\cos{t})e^{t\sin{t}}v, (2t - 2\pi)v , -\frac{1}{2}\sin({\frac{t}{2}})v)$ i.e. 
$$ df_t(v)=((\sin t+t\cos t)e^{t\sin t}v)\frac{\partial}{\partial x}+((2t-2\pi)v)\frac{\partial}{\partial y}-(\frac{1}{2}\sin (\frac{t}{2})v)\frac{\partial}{\partial z}.$$
Then, 
\begin{align*}
(f^{*}\omega)_t(v) &= \left(e^{t\sin{t}}dx + (t^2 - 2\pi t)dy + \cos{\frac{t}{2}}dz\right)df_t(v)\\
&=\left(e^{t\sin{t}}dx + (t^2 - 2\pi t)dy + \cos{\frac{t}{2}}dz\right)\\
&\left(((\sin t+t\cos t)e^{t\sin t}v)\frac{\partial}{\partial x}+((2t-2\pi)v)\frac{\partial}{\partial y}-(\frac{1}{2}\sin (\frac{t}{2})v)\frac{\partial}{\partial z}\right)\\
&=(\sin t+t\cos t)e^{2t\sin t}v+(2t-2\pi)(t^2-2\pi t)v-\frac{1}{2}\cos(\frac{t}{2})\sin(\frac{t}{2})v.
\end{align*}
Note that this unsightly expression is in fact a scalar. This is good, because our differential form evaluated at $t$ is a linear functional $(f^*\omega)_t\in (T_t\mathbb{R})^*$.
The moral of the story is the following: when $f:U\to V$ is given as above, to evaluate $(f^*\omega)_p$ on $v\in T_pU$, you push $v$ forward with $df_p$ then evaluate $\omega$ on it:
$$ (f^*\omega)_p(v)=\omega_{f(p)}(df_pv).$$
I apologize for the unsightly formulas. If something seems wrong - computationally speaking or otherwise - please let me know.
A: Right now, $U$ is in $\mathbb{R}^1$, hence the tangent space considered is also in $\mathbb{R}^1$. This means the $v_1,\ldots,v_k$ is actually just $v_1$ and is, in essence, a number. On the other hand the "${f'(x)v}$" should become a $3$-vector because it is in the tangent space of a space in $\mathbb{R}^3$, which is the tangent vector $f'(t)$ you have calculated, scaled by the number $v_1$.
