I would like to show that for any connexion D on $\mu: N \rightarrow M$ (N, M are manifolds), and any point $n$ in $N$ there is a local basis $\{X_i\}$ of vector fields over $\mu$ such that each $X_i$ is parallel at n, that is, $D_t X_i = 0$ for all t in the tangent space $N_n$.
Can someone point me the way for a proof?
Other answers in this site refer to a Riemmanian structure, orthonormal basis, normal coordinates, etc..
Choose a coordinate system $\varphi = (y^1,\dots,y^k)$ around your point $n \in N^k$ such that $n$ corresponds to $(0,\dots,0)$ and choose some basis $X_1|_{\mu(n)}, \dots, X_m|_{\mu(n)} \in T_{\mu(n)}M$. Define
$$ X_i(\varphi^{-1}(y^1,\dots,y^k)) = P_{\varphi^{-1}(sy^1,\dots,sy^k), s = 0, s = 1}(X_i|_{\mu(n)}). $$
In other words, $X_i$ at (the point corresponding to) $(y^1,\dots,y^k)$ is obtained by parallel transporting $X_i|_{\mu(n)}$ radially from $(0,\dots,0)$ to $(y^1,\dots,y^k)$. The $X_i$ are well-defined, smooth and linearly independent because parallel transport is a linear isomorphism and by constructing, they are parallel (usually only) at $n = \varphi^{-1}(0,\dots,0)$.