Understanding the definition of a pullback of a differential $k$-form and applying it in $1-d$ I am having trouble understanding the definition of a pullback of a differential k-form in a basic course in differentiable geometry.

This is the definition I am given. I believe it is easier to use what I hope is an equivalent identity of
$$ (F^*\beta)(x;u_{(1)},\dots,u_{(n)})=\beta(F(x); \ dF(x)u_{(1)},\dots,dF(x)u_{(n)}) $$
I believe that the definition of algebraic k-form is not widely used so here it is below.

Now in the following example you are asked to apply this definition of a pullback in 1-d.


I am struggling to understand how $\beta(F(x); F'(x)e_{(i)})=\beta_j(F(x))(F'(x)e_{(i)})^j$. (I believe there was a typo there.
I get that 
$\begin{align} \alpha(x;e_{(i)}) &= F^*\beta(x;e_{(i)}) \\
&= \beta(F(x);dF(x)e_{(i)})
\end{align}$ 
Now as $\beta(y)=\beta_j(y)dy^j$ we could either do $y=F(x)$ and then get $\beta_j(F(x))d(F(x))^j$ which would be only half right. Not sure why this doesnt work.
or
$\begin{align} \beta(F(x);dF(x)e_{(i)}) &= (\beta_j(F(x))dy^j)(dF(x)e_{(i)}) \\
&= (\beta_j(F(x))dy^j)(dF(x)e_{(i)}) \\
&= (\beta_j(F(x))d(y^j(dF(x)e_{(i)}) \\
&= (\beta_j(F(x))d(dF(x)e_{(i)}))^j
\end{align}$
Again this is wrong but I am not sure how or why.
Also is $\displaystyle \alpha=F^*\beta=\frac{\partial F^j}{\partial x^i}\beta_j \circ F dx^i$ equivalent to $\displaystyle \alpha=F^*\beta= \beta_j \circ F \frac{\partial F^j}{\partial x^i} dx^i$
Please keep answers simple as this is only a basic course. I am not familiar with tensors or tangent-anything.
 A: Seeing as you seem to use the same book for a lot of your questions, I think the modern/standard approach with tangent vectors will probably just serve to confuse, even though it is easier.
The important point is simply that $\beta$ is a linear form. Assuming I'm understanding your notation correctly, $\beta_{j}(F(x))=\beta(F(x);e_{(j)})$, so $\sum_{j}\beta_j(F(x))(F'(x)e_{(i)})^j=\beta(F(x);F'(x)e_{(i)})$.
Think of $\beta(F(x))$ as a row vector, and $F'(x)e_{(i)}$ as a column vector. Then, this is the same as saying $$\begin{bmatrix}\beta_1(F(x)) & \cdots & \beta_n(F(x))\end{bmatrix}\begin{bmatrix}(F'(x)e_{(i)})^1 \\ \vdots \\ (F'(x)e_{(i)})^n\end{bmatrix}=\beta(F(x);F'(x)e_{(i)}),$$ which is essential true by the definition you gave.
As to your equivalency question, I don't see why multiplication in the reals shouldn't commute. Do you?
A: This is my take on it:
We start with a linear map between vector spaces, to set up the context. 
The pullback of a differential k-form $\theta $, associated to a differentiable mapping $f$ living in the space of $k-$forms  is equivalent to the dual of a multi-linear map between vector spaces $V,W$. The linear map that will be used is the differential $df$   
Let
$V,W$ be finite-dimensional vector spaces, and let $L: V \rightarrow W $ be a linear map (this generalizes to multilinear map). Then the dual map $L^{*}: W^{*} \rightarrow V^{*} $ (notice the "contravariance" of the associated map $L^{*}$) defined by $L^{*}(t)(v_1):= t(L(v_1)); t \in W^{*}$ Basically, we can turn an element in the dual of $W$ into an element on the dual of $V$ using this formula. Now, apply this idea with a map $f: M \rightarrow N$ , both manifolds $V,W$ are now the tangent spaces $T_pM, T_{f(p)}M $  and the linear map $L$ is the differential/pushforward map $df $. Then, given a linear map $L: T_p N \rightarrow  \mathbb R$, i.e., an element of the dual , i.e., a k-form on $T_pN$, you get a k-form defined on $T_pM$ as above
