Let $(M, g)$ be a Riemannian manifold with Levi Civita connection $\nabla: \Gamma(TM) \rightarrow \Gamma( T^*M \otimes_\mathbb{R} TM)$.
I know from here how $\nabla$ induces a map $\Gamma(T^{p,q} M) \rightarrow \Gamma(T^* \otimes T^{p,q} M)$. In particular, I know how to define $\nabla g$.
I've been trying to adapt this to define $\nabla(X) (g)$, where for a fixed $X \in \Gamma(TM)$, by definition $\nabla(X) \in \Gamma( T^*M \otimes_\mathbb{R} TM)$. While I obtained a definition, I'm not sure it's the right one, and I'm struggling to work with it to prove the claim that if we suppose that $Lie_X g = g$, we obtain that $\nabla (X) g = g$ (also using that $\nabla_X g = 0$).
Here's my attempted definition
Denote $E = TM$ as a bundle and $A = \nabla(X)$. So $A: \Gamma(E) \rightarrow \Gamma(E)$. Since $g: \Gamma(TM \otimes TM) \rightarrow \mathcal{C}^\infty(M)$ is equivalent to a tensor in $\Gamma(T^*M \otimes T^*M)$, what I need basically is to be able to extend $A$ to an operator $A^{**}: \Gamma(E^* \otimes E^*) \rightarrow \Gamma(E^* \otimes E^*)$.
Since $\Gamma( \Lambda^1(M) \otimes_\mathbb{R} TM) \simeq \Gamma(T^*M) \otimes_{\mathcal{C}^\infty(M)} \Gamma(TM)$ as $\mathcal{C}^\infty$-modules, and forms are point operators, I can see $\nabla(X)$ as a map $\Gamma(TM) \rightarrow \Gamma(TM)$.
First I define $A^*: \Gamma(E^*) \rightarrow \Gamma(E^*)$ as follows: for $\alpha \in \Gamma(E^*)$ and $s \in \Gamma(E)$, $A^*(\alpha)(s) + \alpha(A(s)) = X (\alpha(s))$.
Then $A^{**}: \Gamma(E^* \otimes E^*) \rightarrow \Gamma(E^* \otimes E^*)$ I define as: $A^{**}( \alpha \otimes \beta) = A^*(\alpha) \otimes \beta + \alpha \otimes A^*(\beta)$ , taking into account that $\Gamma(E^* \otimes E^*) = \Gamma(E^*) \otimes \Gamma(E^*)$.
While this seems to be a good definition, I have no idea how to work with it, because $\nabla(X)^{**}$ is only defined on tensor monomials, but I have no canonical way of writing $g$ as a sum of tensor monomials.
How would the claim that if $Lie_X g = g$, then $\nabla (X) g = g$ be proved starting from this definition (if this definition is good at all)?