Consider the three differentiable functions $\alpha,\beta,h: \mathbb{R}^2 \to \mathbb{R}$ and the associated 1-forms $d\alpha, d\beta$, with $d$ being the exterior derivative.
Let $*$ be the Hodge star operator. Then, is the following relation true:
$$*(h d\alpha \wedge d\beta) = h *(d\alpha \wedge d\beta),$$
namely that the 0-form $h$ can be swapped with the Hodge star operator?
If so, how can one prove it in general?
Yes, the relation you wrote is true. The Hodge star on any manifold $M$ is a pointwise operator and is $\mathbb{R}$-linear at each point $p$ (i.e., $\ast_p: \Lambda^{\cdot} T_p^\ast M \to \Lambda^{\cdot} T_p^\ast M$ is a linear map between vector spaces). This means that for any differential form $\omega$ and any smooth function $h$, at each point $p$, $$(\ast(h \omega))(p) = \ast_p (h(p) \omega(p)) = h(p) \ast_p(\omega(p)) = h(p)\ast(\omega) (p).$$ (I hope my notation is clear.) Thus $\ast(h\omega) = h \ast(\omega)$.
Said in fancy terms, this property of $\ast$ means that it's a vector bundle homomorphism, or a $C^\infty(M)$-module homomorphism.
