The following formula is known among the experts but hard to find in the literature, so I figure I will document it here. Throughout $(M,g)$ denote an arbitrary *pseudo-Riemannian* manifold, and $\nabla$ its Levi-Civita connection.

**Definition** Given a vector field $X$, its corresponding *0th order deformation tensor* is defined to be ${}^{(X,0)}\pi := \mathcal{L}_X g$, where $\mathcal{L}_X$ is Lie differentiation with respect to $X$.

The corresponding *1st order deformation tensor* is defined using a formula similar to that of Christoffel symbols:
$$ {}^{(X,1)}\pi_{ab}{}^c := \frac12 g^{cd} \left[ \nabla_a ( {}^{(X,0)}\pi_{bd}) + \nabla_b ({}^{(X,0)}\pi_{ad}) + \nabla_d ({}^{(X,0)}\pi_{ab}) \right] $$

**Lemma** Let $\Xi$ be an arbitrary $k$-covariant tensor field. And let $X$ be a vector field. The following formula holds for the commutation:
$$ [ \nabla_a, \mathcal{L}_X ] \Xi_{b_1\cdots b_k} = \sum_{j = 1}^k {}^{(X,1)}\pi_{a b_j}{}^c \Xi_{b_1 \cdots b_{j-1} c b_{j+1} \cdots b_k} $$

With the aid of these formulas, we have immediately that, writing $\triangle_g$ for the Laplace-Beltrami operator, first

$$ [ \nabla_X, \triangle_g] f = [\mathcal{L}_X, \triangle_g ] f $$

because Lie derivation and covariant differentiation act identically on scalars, and then

$$ [\mathcal{L}_X, g^{ab}\nabla_a\nabla_b] f = \mathcal{L}_X (g^{ab}) \nabla^a \nabla_b f + g^{ab} [\mathcal{L}_X, \nabla_a] \nabla_b f + g^{ab} \nabla_a [\mathcal{L}_X, \nabla_b ]f $$

The first factor we can compute to get

$$ \mathcal{L}_X(g^{ab}) = - {}^{(X,0)}\pi^{ab} $$

using that $g^{ab} g_{bc} = \delta^a_c$. The third factor vanishes because Lie differentiation commutes with exterior differentiation. And we use our Lemma for the second term. We get, finally

$$ [\nabla_X, \triangle_g] f = - {}^{(X,0)}\pi^{ab} \nabla_a\nabla_b f - g^{ab} ~{}^{(X,1)}\pi_{ab}{}^c \nabla_c f. $$

Remarks:

Notice that the first order deformation tensor is defined in terms of the 0th order one. So that when $X$ is Killing, automatically both the ${}^{(X,0)}\pi$ and ${}^{(X,1)}\pi$ vanish, and differentiation with $X$ commutes with the Laplacian.

In the case ${}^{(X,0)}\pi = \phi g$ for some scalar function $\phi$ (so $X$ is conformally Killing), one can check that the formula reduces to the one I gave in a comment above.

When the function $\phi$ in the previous item is a non-zero constant (which some people refer to as $X$ being a *homothetic vector field*) one gets the special case
$$ [ \nabla_X, \triangle_g] f = \phi \triangle_g f $$