On the wedge product of forms

I suppose I will answer my own question since I think I've figured it out. It comes down to the fact that the functions $a_I$ and $b_J$ are technically 0-forms. Then we use algebraic properties of the wedge product and look at what those algebraic properties tell us when a 0-form is involved.

First, by the anticommutativity of $\wedge$, which says that for a $k$-form $f$ and an $\ell$-form $g$ $$f\wedge g=(-1)^{k\ell}g\wedge f.$$ So, for a 0-form $f$ and an $\ell$-form $g$, we get $f\wedge g=(-1)^{0\cdot \ell} g\wedge f=g\wedge f$.

Second, Tu remarks on page 37 that the wedge product of a 0-form $f$ and an $\ell$-form $\omega$ is actually regular multiplication; that is $f\wedge \omega=f\omega$. So, at a point $p$ we have $(f\wedge\omega)_p=f(p)\omega_p$.

Therefore, if we assume for a moment that $\omega=a_Idx^I$ and $\tau=b_J dx^J$, then we can view this as $\omega=a_I\wedge dx^I$ and $\tau=b_J\wedge dx^J$. Then $$ \omega\wedge \tau=a_I\wedge dx^I\wedge b_J\wedge dx^J=a_I\wedge b_J\wedge dx^I\wedge dx^J=(a_Ib_J)\wedge dx^I\wedge dx^J=(a_Ib_J) dx^I\wedge dx^J. $$ Here's an example: On Problem 4.3 at the end of the section, we are asked to compute $dx\wedge dy$ where $x=r\cos\theta$ and $y=r\sin\theta$. We get: $$ dx=\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \theta}d\theta \\ dy=\frac{\partial y}{\partial r}dr+\frac{\partial y}{\partial \theta}d\theta, $$ and so $$ dx=\cos\theta dr-r\sin\theta d\theta, \\ dy=sin\theta dr+r\cos\theta d\theta. $$ Remember that there is a $\wedge$ between a 0-form and a 1-form; i.e., $\frac{\partial x}{\partial r}dr=\frac{\partial x}{\partial r}\wedge dr$. Now, \begin{align*} dx\wedge dy&= \left(\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \theta}d\theta\right)\wedge\left(\frac{\partial y}{\partial r}dr+\frac{\partial y}{\partial \theta}d\theta\right)\\ &= \left(\frac{\partial x}{\partial r}dr\wedge \frac{\partial y}{\partial r}dr\right) +\left(\frac{\partial x}{\partial r}dr\wedge \frac{\partial y}{\partial\theta}d\theta\right) +\left(\frac{\partial x}{\partial\theta}d\theta\wedge\frac{\partial y}{\partial r}dr\right) +\left(\frac{\partial x}{\partial \theta}d\theta\wedge \frac{\partial y}{\partial\theta}d\theta\right)\\ &= \left(\frac{\partial x}{\partial r} \frac{\partial y}{\partial r}dr\wedge dr\right) +\left(\frac{\partial x}{\partial r} \frac{\partial y}{\partial\theta}dr\wedge d\theta\right) +\left(\frac{\partial x}{\partial\theta}\frac{\partial y}{\partial r}d\theta\wedge dr\right) +\left(\frac{\partial x}{\partial \theta} \frac{\partial y}{\partial\theta}d\theta\wedge d\theta\right)\\ &=0+\left(\frac{\partial x}{\partial r} \frac{\partial y}{\partial\theta}dr\wedge d\theta\right) +\left(\frac{\partial x}{\partial\theta}\frac{\partial y}{\partial r}d\theta\wedge dr\right)+0\\ &=r dr\wedge d\theta. \end{align*} The first equal sign is by definition; the second is because $\wedge$ is distributive over addition; the third is by the anticommutativity of $\wedge$, where, for example we have $dr\wedge \frac{\partial y}{\partial r}=\frac{\partial y}{\partial r}\wedge dr=\frac{\partial y}{\partial r}dr$ in the first parenthesis; the fourth equal sign is by the fact that $dr\wedge dr=0=d\theta\wedge d\theta$; the fifth equal sign follows after plugging in the partials and simplifying.