First, let's remind ourselves of the action of a tensor product. For any $\omega \in \Lambda^p(V^\star), \theta \in \Lambda^q(V^\star)$, we have
$$ (\omega \otimes \theta)(v_1, \dots, v_p, v_{p+1}, \dots, v_{p+q}) = \omega(v_1, \dots, v_p) \times \theta(v_{p+1}, \dots, v_{p+q}).$$
Let's also recall the definition of the $\text{Alt}$ operator. For any $\lambda \in \Lambda^n(V^\star)$, we have
$$ \text{Alt}(\lambda)(v_1, \dots, v_n) = \frac{1}{n!}\sum_{\sigma \in S_n}\text{sgn}(\sigma)\lambda(v_{\sigma(1)}, \dots, v_{\sigma(n)}). $$
The wedge product is constructed by applying the $\text{Alt}$ operator to the tensor product, i.e.
$$ \omega \wedge\theta = \frac{(p+q)!}{p!q!}\text{Alt}(\omega \otimes \theta).$$
So
$$ (\omega \wedge \theta)(v_1, \dots, v_p, v_{p+1}, \dots, v_{p+q}) = \frac{1}{p!q!} \sum_{\sigma \in S_{p+q}} \text{sgn} (\sigma) \omega(v_{\sigma(1)}, \dots, v_{\sigma(p)}) \theta(v_{\sigma(p+1)}, \dots, v_{\sigma(p+q)}). $$
Thus $f^\star(\omega \wedge \theta)(v_1, \dots, v_p, v_{p+1}, \dots, v_{p+q}) = (f^\star\omega \wedge f^\star\theta)(v_1, \dots, v_p, v_{p+1}, \dots, v_{p+q})$, since both of these quantities are equal to
$$ \frac{1}{p!q!} \sum_{\sigma \in S_{p+q}} \text{sgn} (\sigma) \omega(f_\star (v_{\sigma(1)}), \dots, f_\star(v_{\sigma(p)})) \theta(f_\star(v_{\sigma(p+1)}), \dots, f_\star(v_{\sigma(p+q)})). $$
Edit: I notice you've added an extra question at the bottom of your post about $dx \wedge dy \left(\frac{\partial}{\partial y}, \frac{\partial}{\partial x} \right)$. This is a great example for illustration, so let's work through it. The calculation is:
\begin{align*}
dx \wedge dy \left(\tfrac{\partial}{\partial y}, \tfrac{\partial}{\partial x} \right) & = \frac{1}{1!\times 1!} \times \left( dx \left(\tfrac{\partial}{\partial y}\right) dy \left( \tfrac{\partial}{\partial x} \right) - dx \left(\tfrac{\partial}{\partial x}\right) dy \left(\tfrac{\partial}{\partial y} \right) \right) \\
& = 1 \times (0 - 1) \\
& = -1
\end{align*}
