My proof has a hole there, wonder if anyone can help me fill it in?
$$f^*(w \wedge \theta) = (f^*w) \wedge (f^* \theta)$$
By definition,
$$f^*w(x) = (df_x)^*w[f(x)].$$
So I understand $w[f(x)]$ as the $p$-tensor eats a $p$-vector $f(x)$.
The main problem is, that I don't think the function $f$ whose range is a $p$-dimensional vector space can distribute into $w$ and $\theta$ respectively, as what I hope to reach after the question marks.
In order to make $w \otimes \theta$ makes sense on the left hand side, we want $w$ to eat an $m$-vector and $\theta$ to eat an $n$-vector, where $m+n=p$. Hence, how can each $w$ and $\theta$ eats a $p$-vector $f$?
\begin{eqnarray} f^*(w \wedge \theta) & = &(df_x)^* (w \wedge \theta) (f)\\ & = &(df_x)^* \operatorname{Alt}(w \otimes \theta) (f)\\ & = &(df_x)^* \frac{1}{p!}\sum_{\pi \in S_p}(-1)^\pi(w \otimes \theta)^\pi (f)\\ & = & ???\\ & = &\frac{1}{p!}\sum_{\pi \in S_p}(-1)^\pi((df_x)^*w(f) \otimes (df_x)^*\theta(f))^\pi\\ & = & \operatorname{Alt}[(df_x)^*w(f) \otimes (df_x)^*\theta(f)]\\ & = &(df_x)^*w(f) \wedge (df_x)^*\theta(f)\\ & = &(f^*w) \wedge (f^* \theta). \end{eqnarray}
Suppose $\omega$ and $\phi$ are one-forms on $N$ and $f: M \rightarrow N$ is a function. The form $\omega \wedge \phi$ is a two-form on $N$ and the pull-back under $f$ would be denoted $f^*(\omega \wedge \phi)$ it is a two-form on $M$. In particular, the definition can be made precise in terms of two vectors at $p \in M$. \begin{align} f^*(\omega \wedge \phi)(v,w) &= (\omega \wedge \phi) (d_pf(v),d_p(w)) \\ &= (\omega \otimes \phi- \phi \otimes \omega) (d_pf(v),d_p(w)) \\ &= (\omega \otimes \phi)(d_pf(v),d_p(w))- (\phi \otimes \omega) (d_pf(v),d_p(w)) \\ &= \omega (d_pf(v) \phi(d_p(w))- \phi (d_pf(v))\omega (d_p(w)) \\ &= (f^*\omega)(v) (f^*\phi)(w)- (f^*\phi)(v)(f^*\omega)(w) \\ &= (f^*\omega \otimes f^*\phi)(v,w)- (f^*\phi \otimes f^*\omega)(v,w) \\ &= (f^*\omega \otimes f^*\phi - f^*\phi \otimes f^*\omega)(v,w) \\ &= (f^*\omega \wedge f^*\phi)(v,w) \end{align} This holds for all $v,w \in T_pM$ hence the identity $f^*(\omega \wedge \phi) = f^*\omega \wedge f^*\phi$ holds. This argument can be generalized for $\omega$ a $k$-form and $\phi$ a $r$-form, you just have to sort through the antisymmetrization. Looks like you may already have those tools ready to use, so I hope this case suffices to get you started.