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!
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 $\nabla_Xs$.
Best Answer
I'm not sure either, but let me share my thoughts!
As you pointed out correctly, $s^*(s) \in C^{\infty}(M)$, so I suppose $d(s^*(s))$ refers to the exterior differential, hence $d(s^*(s)) \in \Omega^1(M)$. Indeed, we would want to have $\nabla^* s^* \in \Gamma(T^*M \otimes E^*)$ and thus I'd interpret $(\nabla^* s^*)$ as follows: Let $x \in M$ be an arbitrary point. Then $ (\nabla^* s^*)(x) \in (T^*M)_x \otimes (E^*)_x$, and so $(\nabla^* s^*)(x)(s(x)) \in (T^*M)_x = T^*_xM$, for if we write $\nabla^*s^* = \omega \otimes e,$ where $\omega \in \Gamma(T^*M)$ and $e \in \Gamma(E^*)$, we might set $(\nabla^*s^*)s := \omega \cdot (e(s))$, where the dot is just the scalar multiplikation. Thus, $(\nabla^* s^*)(s)$ is indeed a 1 form, as should be. In the same manner, I'd interpret the latter expression $s^*(\nabla s)$, but maybe it's a good exercise for you to write that down in detail by yourself(?).