Change of variables with differential forms In Baby Rudin's theorem 10.22, the proof requires that for $$d(\omega_T)=(d\omega)_T$$ we need $\omega \in \mathcal C'$ and a transformation $T\in \mathcal C"$ where the transform of the k-form $\omega=\sum_I \mathbf{b}(\mathbf{x})dy_I$ ($dy_I$ a basic k-form) is denoted and defined as $$\omega_T=\sum_I \mathbf{b}(T(\mathbf{x}))dt_I$$ and the $t$'s are the components of the transformation. My question is, why is this so?

The proof he gives relies on proving that the result is true for a 0-form $f$ (i.e. $d(f_T)=d(f_T)$).  Then he uses a previously proved result that says that if the k-form $g\in \mathcal C"$ then $d^2g=0$ to show that if $dy_I=dy_{i_1}\wedge\cdots \wedge dy_{i_k}$ then $(dy_I)_T=dt_{i_1}\wedge\cdots \wedge dt_{i_k}$ and therefore that $d((dy_I)_T)=0$ since $T\in \mathcal C"$. 

Here I do not see the need to invoke the previous result, is it not sufficient to use the definition of the derivative of a k-form to show that $$d((dy_I)_T)=(d1)\wedge dt_{i_1}\wedge\cdots \wedge dt_{i_k}=0$$?
And why would that previous result be applicable here since $T$ is not a $k$-form?

Rudin then goes on to say that if $\omega =fdy_I$ then  $\omega_T=f_T(dy_I)_T$ and that to obtain $d(\omega_T)=d(f_T)\wedge (dy_I)_T$ we need the "wedge product rule" derived before. But why does this not follow directly from the definition of the derivative of differential forms? There is no wedge product in $\omega_T=f_T(dy_I)_T$ so that theorem should also not be applicable. What am I missing?
 A: It is likely that this is a nonquestion now, but for future reference here is an answer:
First let us recall the definitions & notations Rudin is using (modified by myself):
Let $E\subseteq \mathbb{R}^n, V\subseteq\mathbb{R}^m$ be open sets, $T\in C^1(E,V)$ ( = Rudin's $\mathcal{C}'$). Let us denote the space of degree $k\;\;C^r$ differential forms on $V$ by
$$\operatorname{Form}^{k,r}(V).$$
Using $x=(x_1,...,x_n)$ ($y=(y_1,...,y_m)$, resp.) for a generic point in $E$ ($V$, resp.), and $t_1,...,t_m$ for the component functions of $T$ we may write
$$T(x)=(t_1(x),...,t_m(x))=(y_1,...,y_m).$$
We may also regard the component functions $t_1,...,t_m$ of $T$ as functions in $C^1(E,\mathbb{R})\cong \operatorname{Form}^{0,1}(E)$, whence their differentials $dt_1,...,dt_m$ are in $\operatorname{Form}^{1,0}(E)$ with
$$dt_i = \sum_{j=1}^n \dfrac{\partial t_i}{\partial x_j} dx_j.$$
In particular $dt_1,...,dt_m$ need not be basic forms, and their coefficient functions may vary depending on the basepoint. Observe that at this point we can't deduce that $ddt_i=0$, since the $\dfrac{\partial t_i}{\partial x_j}$'s are only guaranteed to be continuous (hence the theorem you cited that says $dd=0$ on $\operatorname{Form}^{k,r}(E)$ for $r\geq 2$ does not apply).
Now $T\in C^1(E,V)$ induces a map on the level of form spaces going backwards, called the pullback:
$$T^\ast: \operatorname{Form}^{k,r}(V)\to \operatorname{Form}^{k,0}(E), \\
\left[\omega=\sum_{1\leq i_1<\cdots<i_k\leq m} b_{i_1,...,i_k}\; dy_{i_1}\wedge \cdots\wedge dy_{i_k}\right]  \mapsto \left[\omega_T=\sum_{1\leq i_1<\cdots<i_k\leq m} b_{i_1,...,i_k}\circ T\;\; dt_{i_1}\wedge \cdots\wedge dt_{i_k} \right]. $$
Observe that since the $dt_{i_j}$'s will contibute continuous factors to the coefficient functions (and products of functions are as good as their worst factor) no matter how regular of a form space we start with, the end result is always the space of continuous forms. By the same reasoning we have for $s\geq 1$:
$$C^s(E,V)\times \operatorname{Form}^{k,r}(V)\to \operatorname{Form}^{k,\min{\{r,s-1\}}}(E),\quad (T,\omega)\mapsto T^\ast(\omega).$$
The theorem in question is about the fact that assuming high enough regularity, pullbacks and exterior derivatives commute. More precisely,
Theorem (Baby Rudin, p. 263, 10.22.c):
$$\forall \omega\in \operatorname{Form}^{k,1}(V),\forall T\in C^2(E,V): d \circ T^\ast (\omega) = T^\ast\circ d (\omega)\in \operatorname{Form}^{k+1,0}(E).$$
(It'll be instructive to draw the associated diagram.)
Thus "high enough regularity" above means "as long as all expressions are syntactic in the classical sense (as opposed to the distributional sense)". Observe that the regularity drop is solely due to the exterior derivative.

In light of these remarks, I'll try to answer two of your questions. First there is a problem with
$$d((dy_I)_T)=(d1)\wedge dt_{i_1}\wedge\cdots \wedge dt_{i_k}=0;$$
the $ dt_{i_j}$'s ought to be treated as essentially arbitrary $1$-forms (of $C^1$ regularity, by the assumption that $T\in C^2$), so (for I=(i_1,...,i_k)):
$$d((dy_I)_T)
=d(dt_{i_1}\wedge\cdots\wedge dt_{i_k})\\ 
= (ddt_{i_1})\wedge dt_{i_2}\wedge\cdots\wedge dt_{i_k} - dt_{i_1}\wedge d(dt_{i_2}\wedge\cdots\wedge dt_{i_k}) \\
= - dt_{i_1}\wedge d(dt_{i_2}\wedge\cdots\wedge dt_{i_k})\\
= - dt_{i_1}\wedge (ddt_{i_2})\wedge\cdots\wedge dt_{i_k} + dt_{i_1}\wedge dt_{i_2}\wedge d(dt_{i_3}\wedge\cdots\wedge dt_{i_k})\\
= dt_{i_1}\wedge dt_{i_2}\wedge d(dt_{i_3}\wedge\cdots\wedge dt_{i_k})
= \cdots = 0.$$
Second, I agree that referring to the fact that wedge products and pullbacks commute is a bit verbose, but it is not false: the expression $f(x)dx_1\wedge\cdots dx_k$ can be interpreted as a $k$-form with $f$ as the coefficient function or it can be interpreted as the wedge product of the $0$-form $f$ and the $k$-form $dx_1\wedge\cdots dx_k$. In fact, I submit that it is this hyper-compatibility of the formalism of differential forms that makes them very powerful.
