Problem Statement:
Let $(M,g)$ be a closed (semi-)Riemannian manifold of dimension n and signature (number of negative eigenvalues) $d$. We define the Hodge star operation on it as the unique isomorphism
\begin{align*} \star:\Omega^k(M) &\to \Omega^{n-k}(M), \\ \omega &\mapsto \star \omega \\ \end{align*}
such that the following holds:
$$\forall \omega, \eta \in \Omega^k(M): \omega \wedge \star \eta = (-1)^d ⟨\omega,\eta⟩ \ vol \tag1$$
where $vol := \sqrt{g} \ \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n$ is the canonical volume form, and $⟨ \, ,\, ⟩: \Omega^k(M) \times \Omega^k(M) \to \mathbb{R}$ is the unique extension of the metric to spaces of $k$-forms, defined by
$$⟨v^1\wedge...\wedge v^k,w^1\wedge...\wedge w^k⟩ := \mathrm{det}(g(v^i,w^j)) \tag2$$
where $v^1,...,v^k,w^1,...,w^k \in \Omega(M)$.
It is easy to check that the inverse of the Hodge star operator $\star^{-1} : \Omega^{n-k}(M) \to \Omega^k(M)$ is given by
$$ \star^{-1} = (-1)^{k(n-k)+d}\ \star \tag3$$
Having established that, define the co-differential operator $\delta : \Omega^k(M) \to \Omega^{k-1}(M)$ by the following expression:
$$ \delta := (-1)^k \star^{-1} \circ \ \textrm{d} \circ \star \tag4$$
Now we know the following theorem
$$ \forall \alpha \in \Omega^{k-1}(M) \text{ and } \beta \in \Omega^{k}(M): \quad ⟨ \mathrm{d}\alpha, \beta⟩ = ⟨\alpha, \delta\beta⟩ \tag5$$
Given this setup, let $X \in \mathfrak{X}(M)$ and $ \alpha = \iota_X (g)$. I want to prove that
$$ \mathcal{L}_X (vol) = -\delta \alpha \ vol \tag6$$
Attempt at Solution:
Take an arbitrary $f \in \Omega^0(M)$. Using eqn. (5), we can write
$$ ⟨ \delta \alpha, f⟩ = ⟨ \alpha, \mathrm{d}f⟩ = ⟨ X, grad(f)⟩ = X(f) = \mathcal{L}_X (f) \tag7$$
Now assuming that $ \mathcal{L}_X (fvol) = \mathrm{d}\iota_X (f vol) = 0$, by the Leibniz rule, we have that
$$ \mathcal{L}_X (f) vol = - f \mathcal{L}_X (vol) = -⟨ \mathcal{L}_X (vol), f⟩ \tag8$$
From eqn. (7) and (8), we have our result (6) because $f$ is arbitrary.
Now, the real problem here is to prove that $$\mathrm{d}\iota_X (f vol) = 0 \tag9$$
Since M is closed, I only know that $\int_M \mathrm{d}\iota_X (f vol) = 0$, which actually prompts me to change definition (1) to introduce an integral symbol on the left hand side, which some textbooks do but I do not want to.