Given a decomposition of a vector space $V \simeq U \oplus W$. Then as taking the exterior algebra preserves coproducts (it is left adjoint to the forgetful functor from graded-commutative graded algebras over the base field to vector spaces, taking an algebra to its degree 1 part) one has $$\bigwedge V \simeq \bigwedge U \otimes \bigwedge W \simeq \bigoplus_k \bigoplus_{i + j = k} \bigwedge\nolimits^i U \otimes \bigwedge\nolimits^j W$$ where the multiplication is induced from $(u_1 \otimes w_1) \wedge (u_2 \otimes w_2) = (-1)^{\text{deg}(u_2) \text{deg}(w_1)}(u_1 \wedge u_2) \otimes (w_1 \wedge w_2)$.
Now if one equips $U, W$ with interior products and volume forms $\text{vol}_U$ and $\text{vol}_W$, one obtains Hodge operators $\star_U$ and $\star_W$ on $\bigwedge U$, respectively $\bigwedge W$ by the conditions \begin{align} \langle v, \star u\rangle &= \langle v, u\rangle \cdot \text{vol}_U, \\ \langle v, \star w\rangle &= \langle v, w\rangle \cdot \text{vol}_W. \end{align}
Similarly the interior products induce one on the tensor product and the volume forms induce a volume form $\text{vol} = \text{vol}_U \wedge \text{vol}_W$. Hence there is also a Hodge operator $\ast$ on $\bigwedge V$. Now is it true that $\star = \star_U \otimes \star_W$?