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!
The thing here is that, to establish the second definition, and write a connection on a manifold $M$ as a transformation $\nabla:\Gamma(TM)\rightarrow\nabla(T^*M\otimes TM)$, it is neccesary to use the isomorphism between $\text{Hom}(TM,TM)$ and $T^*M\otimes TM$ in the following way:
Let $\nabla:\mathfrak{X}(M)\times\mathfrak{X}(M)\rightarrow\mathfrak{X}(M)$ a connection on $M$ and $Y\in\mathfrak{X}(M)$. If $p\in M$, we define:
$$\nabla Y(p):T_pM\rightarrow T_pM$$
as $(\nabla Y(p))(v):=(\nabla_X Y)(p)$, with $v\in T_pM$ and $X\in\mathfrak{X}(M)$ any differentiable vector field on $M$ such that $X(p)=v$. So, in particular, $\nabla Y(p)$ will be a linear function, and then:
$$\nabla Y(p)\in\text{Hom}(T_pM,T_pM)$$
which means that $\nabla Y\in\Gamma(\text{Hom}(TM,TM))$. But the bundle $\text{Hom}(TM,TM)$ is isomorphic to $T^*M\otimes TM$. This means that we can consider that:
$$\nabla Y\in\Gamma(T^*M\otimes TM)$$
Based on the above, we can reinterpret a connection on $M$ as a transformation:
$$\nabla:\Gamma(TM)\rightarrow\Gamma(T^*M\otimes TM)$$.
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