It seems to me that people usually include the $\mathbb{R}$-linearity as part of the definition (correct me if I'm wrong); i.e. define $\nabla$ on the simple tensors by the product rule above, and then extend it linearly over $\mathbb{R}$ to the whole space $C^{\infty}(E\otimes F)$.
Actually, due to the fact that $C^{\infty}(E\otimes F)\simeq C^{\infty}(E)\otimes_{C^{\infty}(M)}C^{\infty}(F)$ which OP has mentioned in the comment above, it follows that additivity (i.e.
\begin{align}
\nabla_X(s\otimes t+s'\otimes t')=\nabla_X(s\otimes t)+\nabla_X(s'\otimes t')
\end{align}
which is weaker than $\mathbb{R}$-linearity) is sufficient.
With this in mind, here is an alternative proof, which is essentially a direct computation, though perhaps it may look less elegant.
Let $A\in C^{\infty}(E\otimes F)$. Such $A$ can be expressed as a finite sum of simple tensors, though not uniquely in general. So we want to show that if it can be written in the following two ways
\begin{align}
A=\sum_is_i\otimes t_i=\sum_j\tilde{s}_j\otimes\tilde{t}_j
\end{align}
(the number of summands in these two expressions may be different in general), then
\begin{align}
\nabla_X\left(\sum_is_i\otimes t_i\right)
=\nabla_X\left(\sum_j\tilde{s}_j\otimes\tilde{t}_j\right) & & (*)
\end{align}
Let $(e_{\alpha})$ and $(\epsilon_{\beta})$ be smooth local frames for $E$ and $F$ respectively. Then one can write
\begin{align}
s_i=s_i^{\alpha}e_{\alpha},\qquad
t_j=t_j^{\beta}\epsilon_{\beta},\qquad
\tilde{s}_i=\tilde{s}_i^{\alpha}e_{\alpha},\qquad
\tilde{t}_j=\tilde{t}_j^{\beta}\epsilon_{\beta}
\end{align}
(Einstein summation convention is assumed.) Then we will have
\begin{align}
\sum_is_i\otimes t_i
=\left(\sum_is_i^{\alpha}t_i^{\beta}\right)e_{\alpha}\otimes\epsilon_{\beta},
\qquad
\sum_j\tilde{s}_j\otimes\tilde{t}_j
=\left(\sum_j\tilde{s}_j^{\alpha}\tilde{t}_j^{\beta}\right)e_{\alpha}\otimes\epsilon_{\beta}
\end{align}
Since $\{e_{\alpha}\otimes\epsilon_{\beta}\}$ is a smooth local frame for $E\otimes F$, by uniqueness of local components we must have
\begin{align}
\sum_is_i^{\alpha}t_i^{\beta}=\sum_j\tilde{s}_j^{\alpha}\tilde{t}_j^{\beta} & & (1)
\end{align}
Taking exterior derivative we then have
\begin{align}
\sum_i\left(t_i^{\beta}ds_i^{\alpha}+s_i^{\alpha}dt_i^{\beta}\right)
=\sum_j\left(\tilde{t}_j^{\beta}d\tilde{s}_j^{\alpha}
+\tilde{s}_j^{\alpha}d\tilde{t}_j^{\beta}\right) & & (2)
\end{align}
Now let $\omega_{\alpha}^{\beta}$ and $\theta_{\alpha}^{\beta}$ be the connection 1-forms of $\nabla^E$ and $\nabla^F$ respectively, so that e.g.
\begin{align}
\nabla_X s_i=\left[Xs_i^{\alpha}+s_i^{\beta}\omega_{\beta}^{\alpha} (X)\right]e_{\alpha} & & (3)
\end{align}
and similar identities hold for $\nabla_Xt_i$, $\nabla_X\tilde{s}_j$ and $\nabla_X\tilde{t}_j$. Then we can compute
\begin{align}
&\nabla_X\left(\sum_is_i\otimes t_i\right) \\
%%%
&=\sum_i\left(\nabla_Xs_i\otimes t_i+s_i\otimes\nabla_Xt_i\right) \\
%%%
&=\underbrace{\left(\sum_i\left((Xs_i^{\alpha})t_i^{\beta}
+s_i^{\alpha}(Xt_i^{\beta})\right)\right)}_{=:I}e_{\alpha}\otimes\epsilon_{\beta}
+\underbrace{\left(\sum_is_i^{\gamma}t_i^{\beta}\right)}_{=:II}
\omega_{\gamma}^{\alpha}(X)e_{\alpha}\otimes\epsilon_{\beta}
+\underbrace{\left(\sum_is_i^{\alpha}t_i^{\gamma}\right)}_{=:III}
\theta_{\gamma}^{\beta}(X)e_{\alpha}\otimes\epsilon_{\beta}
\end{align}
where the first step follows by additivity and product rule, while the second step is obtained by substitution of (3) and rearranging the terms.
Now $\nabla_X\left(\sum_j\tilde{s}_j\otimes\tilde{t}_j\right)$ will have the same expression, except that all of the $s_i,t_i$ are replaced by $\tilde{s}_j,\tilde{t}_j$; says
\begin{align}
\nabla_X\left(\sum_j\tilde{s}_j\otimes\tilde{t}_j\right)
=\tilde{I}\cdot e_{\alpha}\otimes\epsilon_{\beta}
+\tilde{II}\cdot \omega_{\gamma}^{\alpha}(X)e_{\alpha}\otimes\epsilon_{\beta}
+\tilde{III}\cdot \theta_{\gamma}^{\beta}(X)e_{\alpha}\otimes\epsilon_{\beta}
\end{align}
Then:
- By (2), we have $I=\tilde{I}$.
- By (1), we have $II=\tilde{II}$ and $III=\tilde{III}$.
Hence, (*) holds as desired.
Let's work with local frames. Suppose that $\{ e_i \}$ is a local frame for $\xi$, and suppose that $\{ e_j^\star \}$ is the dual frame for $\xi^{\star}$ (i.e. $\langle e_i , e_j^\star \rangle = \delta_{ij}$). And suppose that $\nabla_{\xi} e_i = \sum_k\omega_{ik} e_k$ for suitable one-forms $\omega_{ik}$. An arbitrary section of $\xi^\star$ can be written as $s^\star = \sum_j s^\star_j e_j^\star$, for suitable smooth functions $s_j$, and our aim is to determine $\nabla_{\xi^\star} s$. We can calculate as follows:
\begin{align} (e_i, \nabla_{\xi^\star}s^\star) &= d(e_i, s^\star) - (\nabla_\xi e_i , s^\star) \\ &= \sum_j d\left( s^\star _j \langle e_i, e_j^\star \rangle \right) -\sum_{jk} s^\star_j \omega_{ik}\langle e_k, e_j^\star \rangle \\ &=ds^\star_i-\sum_j \omega_{ij}s^\star_j. \end{align}
This implies that
$$ \nabla_{\xi^\star} s^\star = \sum_i( ds^\star_i - \sum_j \omega_{ij} s^\star_j )e_i^\star. \ \ \ (\star)$$
We still need to show that the $\nabla_{\xi^\star}$ defined in $(\star)$ is a connection. This follows from this calculation:
\begin{align} \nabla_{\xi^\star} (fs^\star) &= \sum_i( d(fs^\star_i) - \sum_j \omega_{ij} fs^\star_j) e^\star_i \\ &=(df) \sum_is^\star_i e^\star_i + f \sum_i ( ds^\star_i - \sum_j \omega_{ij} s^\star_j) e^\star_i \\ &= (df) s^\star + f\nabla_{\xi^\star} s^\star \end{align}
We also need to show that $d(s,s^*) = (\nabla_{\xi}(s), s^*) + (s, \nabla_{\xi^*} (s^*))$ holds for arbitrary $s$ when $\nabla_{\xi^\star}$ is defined by $(\star)$. Writing $s$ in the form $s = \sum_i s_i e_i$ for suitable smooth functions $s_i$, this statement follows from the calculations:\begin{align} d(s, s^\star) &= d\left( \sum_{ij} s_i s^\star_j\langle e_i, e_j^\star \rangle \right) = \sum_i d(s_i s^\star_i) =\sum_i(ds_i) s_i^\star + s_i (ds^\star_i) \\ (\nabla_\xi s, s^\star) &= \sum_{ij} (ds_i + \sum_k \omega_{ki}s_k) s^\star _j\langle e_i, e_j^\star \rangle =\sum_i (ds_i) s_i^\star + \sum_{ik} \omega_{ki} s_k s_i^\star \\ (s, \nabla_{\xi^\star} s^\star ) &= \sum_{ij} s_i (ds_j^\star - \sum_k \omega_{jk} s_k^\star ) \langle e_i , e_j^\star \rangle =\sum_j s_j (ds^\star_j) - \sum_{jk}\omega_{jk} s_j s_k^\star \end{align}
Edit: As for coordinate independence, one way is to do the brute force calculation. If $\{ f_i \}$ is a different local frame, valid on a different patch, with $f_i = \sum_{ij} g_{ij} e_j$ on the overlap between the two patches for suitable non-vanishing smooth functions $g_{ij}$, then the dual bases are related by $f_i^\star = \sum_{ij}(g^{-1})_{ji} e_j^\star$. And if $\nabla_\xi f_i = \sum_k \zeta_{ik} f_k$ for suitable one-forms $\zeta_{ik}$, then you can show that
$$ \zeta_{ik} = \sum_l dg_{il} (g^{-1})_{lk} + \sum_{lm} g_{il} \omega_{lm} (g^{-1})_{mk},$$
and from here, you can verify that definition $(\star)$ is consistent whether you work using the $\{ e_i \}$ frame or the $\{ f_i \}$ frame.
But actually, I don't think we need to do all this work! Since $(\star)$ is the unique $\nabla_{\xi^\star}s^\star$ satisfying $d(s,s^*) = (\nabla_{\xi}(s), s^*) + (s, \nabla_{\xi^*} (s^*))$ for all $s$ on the first patch, and since $(\star)$-written-in-the-new-frame is the unique $\nabla_{\xi^\star}s^\star$ satisfying $d(s,s^*) = (\nabla_{\xi}(s), s^*) + (s, \nabla_{\xi^*} (s^*))$ for all $s$ on the second patch, the two definitions must be equal on the overlap!
Best Answer
Your expression is incorrect. In particular, I'm not sure why you introduced $T$ when there is no $T$ in the expression $\nabla_Xs$.
Here's a slightly different (but completely equivalent) way to think of $\nabla_Xs$. Note that $$\nabla s \in A^1(E) = \Gamma(M, T^*M\otimes E) \cong \Gamma(M, \operatorname{Hom}(TM, E)).$$ If $X \in \Gamma(M, TM)$, then $(\nabla s)(X) \in \Gamma(M, E)$ is denoted $\nabla_Xs$.