The short answer is yes.
The easiest long answer I can think of uses the language of principal bundles and associated vector bundles. If you have not encountered these objects, then I can recommend a few sources for you, but the basics of them isn't too hard to wrap ones head around.
Let $E$ and $F$ be smooth vector bundles over $M$ with covariant derivatives $\nabla^E$ and $\nabla^F$ which are metric compatible with the bundle metrics $\langle \cdot ,\cdot \rangle_E$ and $\langle \cdot, \cdot \rangle_F$. Note that the induced bundle metric on $E\otimes F$ is given pointwise on simple tensors by :
$$\langle \phi_E\otimes \phi_F,\psi_E\otimes \psi_F \rangle_{(E\otimes F)_x}=\langle \phi_E ,\psi_E \rangle_{E_x}\cdot \langle \phi_F,\psi_F \rangle_{F_x}$$
First note that since $E$ and $F$ metrics on them, there exists a bundle of orthonormal frames of $E$ and $F$, which we denote by $O(E)$, and $O(F)$ respectively. If the fibre of $E$ is isomorphic to $\mathbb{R}^n$, and the fibre of $F$ is isomorphic to $\mathbb{R}^m$ we have that $O(E)$ and $O(F)$ admit smooth right actions of $O(n)$ and $O(m)$ respectively, which preserve the fibres of the bundles, and are free and transitive on them.
We can then describe $E$ and $F$ as vector bundles associated these principal bundles. Indeed, if $\rho_n$ and $\rho_m$ denote the standard representations of $O(n)$ and $O(m)$ on $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively, then we have that as vector bundles:
$$E\cong O(E)\times_{\rho_n}\mathbb{R}^n$$
where $O(E)\times_{\rho_n}\mathbb{R}^n$ is the quotient space defined by the equivalence relation:
$$(p,v)\sim (p\cdot g, \rho_n(g)^{-1} v)$$
Writing points in $O(E)$ as a frame $(v_1,\dots, v_n)_p$ for $E_p$, and a vector in $\mathbb{R}^n$ as $(x^1,\dots, x^n)$, the isomorphism is given by:
$$[(v_1,\dots, v_n)_p, (x^1,\cdots, x^n)]\longmapsto v_ix^i$$
It is easy to see that this assignment is independent of our choice of class representative, and checking that this is indeed smooth and a fibre wise vector space isomorphism is also done easily. Similar results hold for $F$.
Let $\langle \cdot ,\cdot \rangle_{\mathbb{R}^n}$ be the standard inner product on $\mathbb{R}^n$, then the original bundle metric $E$ can be rewritten as:
$$\langle [p,v], [p,w]\rangle_{E_x}= \langle v,w\rangle_{\mathbb{R}^n}$$
where $\pi(p)=x$. This should make intuitive sense, as this operation is well defined since $\rho_n$ leaves $\langle \cdot ,\cdot \rangle_{\mathbb{R}^n}$ invariant so this is well defined, and in this framework we are essentially writing everything in terms of orthonormal frames of $E$.
The metric compatible covariant derivative's $\nabla^E$, $\nabla^F$ induce connection one forms $A_E$ and $A_F$ in the principal bundles $O(E)$ and $O(F)$. These are one forms valued in the Lie algebras $\mathfrak{o}(n)$ and $\mathfrak{o}(m)$, and satisfy two properties. Namely, that under pullback by right multiplication we have:
\begin{align*}
R_g^* A_E=Ad_{g^{-1}}\circ A_E
\end{align*}
and for all fundamental vector field $\tilde{X}$ associated to a Lie algebra element $X$ we have:
\begin{align*}
A_E(\tilde{X})=X
\end{align*}
Here, $\tilde{X}$ is defined pointwise by:
\begin{align*}
\tilde{X}_p=\frac{d}{dt}\Big|_{t=0}p\cdot \exp(tX)
\end{align*}
Furthermore, we can reobtain the metric compatible covariant derivative by defining a covariant derivative locally by noting that for a smooth section $s:U\rightarrow O(E)_U$, local sections $\Phi:U\rightarrow E_U$, are in one to correspondence with smooth maps $\phi:U\rightarrow \mathbb{R}^n$. It follows that given a global section $\Phi:M\rightarrow E$, and a local section $s:U\rightarrow O(E)_U$:
$$\Phi|_U=[s, \phi]$$
for some unique smooth map $\phi:U\rightarrow \mathbb{R}^n$. We thus define the covariant derivative locally by:
$$\nabla^E_X\Phi|_U=[s, d\phi(X)+\rho_{n*}(A_{E_s}(X))\phi]$$
where $\rho_{n*}$ is the induces representation of $\mathfrak{o}(n)$ on $\mathbb{R}^n$, and $A_{E_s}$ is the pull back of $A_E$ by $s$. One can check that this is independent of the section we chose, so we this defines a global operation. Furthermore, since:
\begin{align}
\langle \rho_{n*}(X)v, w\rangle_{\mathbb{R}^n}+\langle v, \rho_*(X)w \rangle_{\mathbb{R}^n}=0
\end{align}
for all $X\in \mathfrak{o}(n)$, one can check that
this prescription is metric compatible with the bundle metric $\langle \cdot ,\cdot \rangle_E$. We can do similar things for $F$.
Ok, but what does this tell us about $E\otimes F$? Well it turns out we can also describe $E\otimes F$ as vector bundle associated to some principal bundle: the fibre wise product of $O(E)$ and $O(F)$. Indeed, we define the fibrewise product by:
$$O(E)\times_M O(F)=\coprod_{x\in M} \pi_{O(E)}^{-1}(x)\times \pi_{O(F)}^{-1}(x)$$
As a set, this is equivalent to:
$$\{(p,q)\in O(E)\times O(F): \pi_{O(E)}(p)=\pi_{O(F)}(q)\}$$
This set can be shown to be principal $O(n)\times O(m)$ principal bundle, where for all $(g,h)\in O(n)\times O(m)$, the right action is given by:
$$(p,q)\cdot (g,h)=(p\cdot g,q\cdot h)$$
Furthermore, we obtain a new representation $\rho_n\otimes \rho_m$ on $\mathbb{R}^n\otimes \mathbb{R}^m$ given on simple tensors by:
$$(\rho_n\otimes \rho_m)(g,h)\cdot(v\otimes w)=\rho_n\cdot(g)v\otimes \rho_m(h)\cdot w$$
We then have that as vector bundles:
$$E\otimes F\cong (O(E)\times _M O(F))\times_{\rho_n\otimes \rho_m} (\mathbb{R}^n\otimes \mathbb{R}^m)$$
We have an induced inner product on $\langle\cdot,\cdot \rangle_{\mathbb{R}^n\otimes \mathbb{R}^m}$, given by the standard inner products on $\mathbb{R}^n$ and $\mathbb{R}^M$. Via the same process as before, this induces the the inner product on $E\otimes F$.
The connection one forms on $O(E)$ and $O(F)$ define a new connection one form $A_E\oplus A_F$ on $O(E)\times_M O(F)$ given by:
$$(A_E\oplus A_F)_(p,q)(X_p,Y_q)=(A_E(X_p), A_F(Y_q))$$
for all $(X_p,Y_q)\in T_pO(E)\times T_qO(F)$ where $\pi_{O(E)}(p)=\pi_{O(F)}(q)$, and $\pi_{O(E)*}X_p=\pi_{O(F)*}Y_q$. This connection one form induces the correct covariant derivative on $E\otimes F$. Let's see this; let $\Phi\in \Gamma(E)$ and $\Psi\in\Gamma(F)$, then:
\begin{align*}
\nabla^{E\otimes F}_X(\Phi\otimes \Psi)|_U=&[s_E\times_M s_F, d(\phi\otimes \psi)+(\rho_{n}\otimes \rho_m)_*((A_{E_{s_E}}\oplus A_{F_{s_F}})(X))(\phi\otimes\psi)]\\
=&[s_E\times s_F,d\phi\otimes \psi+\rho_{n*}(A_{E_{s_E}}(X))\phi\otimes \psi]+[s_E\times s_F,\phi\otimes d\psi+\phi\otimes\rho_{m*}(A_{F_{s_F}}(X))\psi]\\
=&\nabla^E_X\Phi\otimes \Psi|_U+\Phi\otimes \nabla^F_X\Psi|_U
\end{align*}
We want to check that this is connection is metric compatible, however this trivially follows from the earlier discussion, as the representation $\rho_n\otimes \rho_m$ leaves the inner product $\langle \cdot,\cdot\rangle_{\mathbb{R}^n\otimes \mathbb{R}^m}$ invariant, so we'll obtain a similar rule for the induced representation $(\rho_n\otimes \rho_m)_*$.
So since $Hom(E,F)\cong E^*\otimes F$, we then just need to show that the induced covariant derivative is on $E^*$ is metric compatible. We will use the frame bundle $O(E)$ to construct $E^*$ as a vector bundle associated to $O(E)$. Let $\mathbb{R}^{n*}$ be the dual space to $\mathbb{R}^n$, then there is an induced representation $\rho'_n$ of $O(n)$ on $\mathbb{R}^n$ given implicitly by:
$$(\rho'_n(g)\cdot \omega)(v)=\omega(\rho_n(g^{-1})v)$$
One can then show by the definition of $E^*$:
$$E^*\cong O(E)\times_{\rho'_n} \mathbb{R}^{n*}$$
Note that the inner product on $\mathbb{R}^{n}$ induces an isomorphism $\mathbb{R}^n\rightarrow \mathbb{R}^{n*}$ given by:
$$f:v\longmapsto \langle v,\cdot\rangle_{\mathbb{R}^n}$$
implying that $f(v)(w)=\langle v,\cdot w\rangle$. The inverse
$f^{-1}$ induces the inner product on $\mathbb{R}^{n*}$ given by:
\begin{align*}
\langle \omega, \eta\rangle_{\mathbb{R}^{n*}}=\langle f^{-1}(\omega),f^{-1}(\eta)\rangle_{\mathbb{R}^n}
\end{align*}
We need to check that $\rho'_n$ leaves this inner product invariant. We have that:
\begin{align*}
\langle \rho'_n(g)\cdot \omega,\rho'_n(g)\cdot \eta \rangle_{\mathbb{R}^{n*}}=&
\langle f^{-1}(\rho'_n(g)\cdot \omega), f^{-1}(\rho'_n(g)\cdot \eta)\rangle_{\mathbb{R}^n}
\end{align*}
Note that we have:
\begin{align*}
f(\rho(g)v)(w)=\langle \rho_n(g)v,w\rangle_{\mathbb{R}^n}=\langle v,\rho_n(g)^Tw\rangle_{\mathbb{R}^n}
\end{align*}
Since $\rho(g)^T=\rho(g)^{-1}$ we obtain that:
$$f(\rho_n(g)v)(w)=f(v)(\rho_n(g^{-1})w)=\rho'_n(g)\cdot f(v)$$
It then follows that for some unique $v\in \mathbb{R}^n$:
\begin{align*}
f^{-1}(\rho'_n(g)\cdot \omega)=&f^{-1}(\rho'_n(g)\cdot f(v))\\
=&f^{-1}(f(\rho_n(g)\cdot v))\\
=&\rho_n(g)\cdot v
\end{align*}
So if $\omega=f(v)$ , and $\eta=f(w)$ we have that:
\begin{align*}
\langle \rho'_n(g)\cdot \omega,\rho'_n(g)\cdot \eta \rangle_{\mathbb{R}^{n*}}=&
\langle f^{-1}(\rho'_n(g)\cdot \omega), f^{-1}(\rho'_n(g)\cdot \eta)\rangle_{\mathbb{R}^n}\\
=&\langle f^{-1}(f(\rho_n(g)\cdot \omega), f^{-1}(f(\rho_n(g)w))\rangle_{\mathbb{R}^n}\\
=&\langle \rho_n(g)v,\rho_n(g)w\rangle_{\mathbb{R}^n}\\
=&\langle v,w\rangle_{\mathbb{R}^n}\\
=&\langle \omega,\eta\rangle_{\mathbb{R}^{n*}}
\end{align*}
This inner product then induces the same inner product on $E^*$ as the one obtained by the bundle isomorphism $F:E\rightarrow E^*$ induced by $\langle \cdot ,\cdot \rangle_{E}$. It is easy to check that the covariant derivative defined in the language of connection one forms is then the induced covariant derivative, and since the $\rho'_n$ leaves the inner product on $\mathbb{R}^{n*}$ invariant, it follows that the induced covariant derivative is also metric compatible.
From our earlier discussion we then have that the induced covariant derivative on $E^*\otimes F$ is metric compatible with the induced metric, as desired. You can check that this covariant derivative will be the same as the one you defined implicitly above.
