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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..

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    What do you mean by "connection on $\mu$"? A connection is usually defined for a vector bundle $E \rightarrow M$ over a manifold $M$. If the vector bundle is $TM$ then we say this is (an affine) connection on $M$. – 2017-02-06
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    An affine connection $D$ on $\mu$ assigns to each element on $N_n$ an operator which maps vector fields over $\mu$ into elements of $M_{\mu (n)}$ that satisfies the usual axioms for connections (linearity over the elements of $N_n$, linearity over the vector fields, Leibniz rule and smoothness). Just in case, a vector field over $\mu$ assigns to each $n$ in $N$ a vector in $M_{\mu (n)}$. I'm learning through an old book, maybe that is the case. I believe the definitions are very similar. – 2017-02-06
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    Ah, ok. This is usually called a connection along $\mu$ or a connection on the pullback bundle $\mu^{*}(TM)$ over $N$. – 2017-02-06

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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 $n = \varphi^{-1}(0,\dots,0)$.