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.
Best Answer
Look at the second paragraph on page 91 of Introduction to Riemannian Manifolds (2nd edition). It says
Proposition 4.3 shows that a connection on a vector bundle $E\to M$ automatically determines a unique connection on the restriction of $E$ to any open subset of $M$; this is what justifies applying the connection to a vector field defined only on an open subset. Proposition 4.5 then says that the value of $\nabla_X Y$ at a point $p$ depends only on the value of $X$ at $p$; this is what justifies the notation $\nabla_v Y$ for a vector $v\in T_p M$.
To actually compute the value of $\nabla_v Y$ at a point $p\in M$, just choose a local frame defined on a neighborhood of $p$, compute the connection coefficients in that local frame, and use the formula you displayed in your post. There's no need to extend anything to a global vector field.