Pullback expanded form. 
Definition. If $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, define a $p$-form $f^*\omega$ on $X$ as follows: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$

According to Daniel Robert-Nicoud's nice answer to $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, what is $\omega[f(x)]$?, locally, differential form can be written as
$$\omega_\alpha(y) = \alpha(y)dx^{i_1}\wedge\ldots\wedge dx^{i_p}$$
with $\alpha$ a smooth function. Then
$$f^*\omega_\alpha(x) = (df_x)^*[(\alpha\circ f)(x)dx^{i_1}\wedge\ldots\wedge dx^{i_p}].$$
Hence, if $$\omega_\alpha(y) = \alpha(y)dx^{i_1}\wedge\ldots\wedge dx^{i_p},$$
$$\theta_\beta(y) = \beta(y)dx^{j_1}\wedge\ldots\wedge dx^{j_q},$$
Do I get $\omega \wedge \theta$ such that
$$\omega_\alpha \wedge \theta_\beta(y)= \gamma(y)dx^{k_1}\wedge\ldots\wedge dx^{k_{p+q}}?$$
If so, how can I prove it?
And can I write $\gamma$ in terms of $\alpha$ and $\beta$?
Thank you~~~

I suppose generally a $p$-form is a sum of such terms, but if we can understand how one such element pulls-back then linearity extends to $\sum_{i_1, \dots , i_p}\alpha_{i_1,\dots , i_p} dy^{i_1} \wedge \cdots \wedge dy^{i_p}$. That said, to calculate $\gamma$ you just have to sort out the sign needed to arrange the indices on the wedge of $\omega_{\alpha} \wedge \theta_{\beta}$. I prefer the notation, assuming $I \cap J  = \emptyset$,
$$ (\alpha dy^I) \wedge (\beta dy^J) = \alpha \beta dy^I\wedge dy^J =  \alpha \beta (-1)^{\sigma(I,J)}dy^K $$
here $\alpha\beta$ merely indicated the product of the scalar-valued functions $\alpha$ and $\beta$ and I have suppressed the $y$-dependence as it has little to do with the question. Here $\sigma(I,J)$ is the number of transpositions needed to rearrange $(I|J)$ into $K$.
Of course, you could just leave the $I$ and $J$ unchanged in which case the $\gamma = \alpha \beta$. I rearranged them because in some of what you are interested in reading there will be a supposition that the indices are arranged in increasing order so if $I = (1,2,5)$ and $J = (3,6)$ then you'll want $(1,2,5)(3,6) \rightarrow K = (1,2,3,5,6)$ which requires flipping $3$ and $5$ hence $\sigma(I,J) = 1$. Ok, usually we use "sgn" of a permutation to get this sign so my notation is a bit nonstandard. 
