For a (compact) Riemannian manifold $(M,g)$, can it happen that for a non-zero form $\text{d}^\*\omega$, and a smooth function $f$ such that $\text{d}f \neq 0$, we can have
$$
\text{d}f \wedge \text{d}^\*\omega = 0?
$$ 
Note that $\*$ denotes the codifferential with respect to $g$. 



Vanishing product of a closed and coclosed form on a Riemannian manifold
Yes, this can happen: Take the flat torus $T=\mathbb R^2/\mathbb Z^2$, $$f\colon T\to \mathbb R; x\mapsto \sin(2\pi x),$$
$$g\colon T\to \mathbb R; y\mapsto \sin(2\pi y),$$
and $$\omega=g \text{vol}= g dx\wedge dy.$$ Then, $$df\wedge d^\* g=\pm \cos(2\pi x)\cos(2\pi y) dx\wedge dx=0,$$
where the actual sign does not matter.


