Let $(M, g)$ be a Riemann manifold and $$\nabla^{LC}: \Gamma(M,TM) \to \Gamma(M, T^*M \otimes TM)$$ the Levi-Civita connection over the tangential bundle $p:TM \to M$.
Since in general for arbitrary two connections $\nabla^E, \nabla^F$ on vector bundles $E \to M, F \to M$ there is a canonical way to form the tensor connection
$\nabla^{E \otimes F}: \Gamma(M,E \otimes F) \to \Gamma(M, T^*M \otimes E \otimes F)$ of the bundle $E \otimes F \to M$ by the rule
$$\nabla^{E \otimes F}(a \otimes b):= (\nabla^E a ) \otimes b+ a \otimes (\nabla^F b)$$
and the dual connection $\nabla^{E^*}: \Gamma(M,E^*) \to \Gamma(M, T^*M \otimes E^*)$ of the dual bundle $E^* \to M$ by the rule
$$(\nabla^{E^*} \alpha)(a)= d(\alpha(a))- \alpha(\nabla^{E} a)$$
for $a \in \Gamma(M,E), \alpha \in \Gamma(M, E^*)$.
we can ask following question:
Since $g \in \Gamma(M, (T^*M)^{\otimes 2})$ the Levi-Civita connection $\nabla^{LC}$ extends as explaned above in canonical way to a connection on bundle $(T^*M)^{\otimes 2} \to M$ and as well to a connection on $\operatorname{End} {E} \to M$ since $\operatorname{End} E= E^* \otimes E$.
Let $A \in \Gamma(M, \operatorname{End} E)$. Therefore it make sense to talk about the object $\nabla^{LC} A$. We define now a new $2$-form
$$h := g(A (-),-) \in \Gamma(M, (T^*M)^{\otimes 2})$$
given on elements $X,Y \in \Gamma(M, TM)$ by
$$h(X,Y)= g(JX,Y)$$
and as before the Levi Civita acts also on $h$.
The Question is why for $(\nabla^{LC} h)(X,Y)$ following rule holds:
$$(\nabla^{LC} h)(X,Y)= (\nabla^{LC} g)(AX,Y)+ g( (\nabla^{LC} A) \cdot X, Y)$$?
Everything boils down to ask why $\nabla^{LC}(A \cdot X)= (\nabla^{LC} A) \cdot X + A \cdot (\nabla^{LC} X)$ but I can't find an answer.
Best Answer
First I wanted to say that a 2-tensor (in $T^*M\otimes T^*M$) is in general not necessarily a "2-form" (i.e. not necessarily in $T^*M\wedge T^*M$).
Anyway, you have all the pieces laid out. In the case that $A$ is a simple tensor $a\otimes \alpha$, with $a\in \Gamma(E),\alpha\in \Gamma(E^*)$, then $$ \begin{gather*} (\nabla A)(X) = (\nabla a \otimes \alpha + a \otimes \nabla\alpha)(X) \\ = (\alpha(X))\nabla a + \big(d (\alpha(X)) - \alpha(\nabla X)\big) a \\ = \big( (\alpha(X))\nabla a + d (\alpha(X))a \big) - \alpha(\nabla X) a \\ = \nabla(A(X)) - A(\nabla X) \end{gather*} $$ (where I suppressed any $\otimes$ coming from the 1-form part of $\nabla$ or $d$ (for example, in the third line it should technically be $d(\alpha(X))\otimes a$).
In general $A$ is just locally a sum of simple tensor (fields), and this is a local statement.
Then $$\begin{gather*} (\nabla h)(X,Y) = d(h(X,Y)) - h(\nabla X, Y) - h(X,\nabla Y) \\ = d(g(JX,Y)) - g(J\nabla X,Y) - g(JX,\nabla Y) \\ = d(g(JX,Y)) - g(\nabla (JX),Y) + g((\nabla J)X,Y) - g(JX,\nabla Y). \end{gather*}$$ Meanwhile $(\nabla g)(JX,Y)$ is $$(\nabla g)(JX,Y) = d(g(JX,Y)) - g(\nabla(JX),Y) - g(JX,\nabla Y),$$ so using these two, we have $$(\nabla h)(X,Y) = (\nabla g)(X,Y) + g((\nabla J)X,Y)$$ as desired, but I think you already did this part.