Let $D$ be an $m$-dimensional distribution on an $n$-dimensional manifold $M$. Let $U\subset M$ be an arbitrary open subset such that on $U$ we can define vector fields $e_i$ such that for each $x\in U$, $e\equiv\{e_1(x),...,e_n(x)\}$ is a basis for $T_xM$ and $\{e_1(x),...,e_m(x)\}$ is a basis for $D_x\subset T_xM$. Then if we have a connection $\nabla:\Gamma(TM)\times\Gamma(TM) \rightarrow \Gamma(TM)$ on $M$, its restriction to $U$ can be written in terms of the connection matrix $\alpha_U$ (a matrix valued 1-form on $U$), defined by $\nabla_X(e_i)=\sum_{j=1}^n\alpha_{U,ij}(X)e_j$.
I write $\alpha_U= \begin{pmatrix} A & B \\ C & D \end{pmatrix}$ for $A\in\mathbb{R}^{m\times m}$, $B\in\mathbb{R}^{m\times (n-m)}$, $C\in\mathbb{R}^{(n-m)\times m}$, $D\in \mathbb{R}^{(n-m)\times (n-m)}$.
I'm trying to prove that for any such $U$ and $e$ we have that $\alpha_U$ is such that $B=0$ if and only if $\nabla_X(Y)\in \Gamma(D)$ for all $X\in\Gamma(TM)$ and for all $Y\in\Gamma(D)$.
My problem is that the expression $\nabla_X(Y)\in \Gamma(D)$ is a global one. For the $\Leftarrow$ direction this doesn't matter: if $\nabla_X(Y)\in \Gamma(D)$ holds then I can use the local expression in terms of $\alpha$ and deduce that $B=0$.
However, for $\Rightarrow$ (so given that $B=0$), I can pick an arbitrary $X\in \Gamma(TM)$ and $Y \in \Gamma(D)$. Then I can deduce that when restricted to an open subset $U$ as above, we indeed get that $\nabla_X(Y)$ is a tangent vector in $D_x$ for each $x\in U$. But how can I derive that $\nabla_X(Y)\in \Gamma(D)$ also holds globally?