Consider a smooth manifold $M$, with a local basis $\{\mathbf e_i\}_{i=1}^n$. Can a connection $\nabla$ such that $\nabla_i\mathbf e_j=0$ for all $i,j$ always be defined?
For context, the question comes from watching some lecture videos on youtube (e.g. this one) explaining tensor calculus notation. In there, the speaker defines the covariant derivative in such a way that $\nabla_i\mathbf e_j=0$ for all $i,j$, so that we have $$\partial_i \mathbf V = (\partial_i V^k + \Gamma^k_{~ij}V^j)\mathbf e_k \equiv (\nabla_i V^k)\mathbf e_k, \qquad \mathbf V\equiv V^j\mathbf e_j,$$ defining $\nabla_i V^k\equiv \partial_i V^k + \Gamma^k_{~ij}V^j$. I think the expression $\nabla_i V^k$ can be understood as simply shorthand for $\mathbf e^k(\nabla_i\mathbf V)$ (as otherwise, what would it mean for a covariant derivative to act on functions rather than vector/tensor fields?), but I am not confident about this.
All this is introduced, in the linked videos, in the context of flat surfaces embedded in $\mathbb R^n$. What I am looking for here is the more abstract justification for these types of formal manipulations. In particular, in the video they end up with a covariant derivative such that $\nabla_i\mathbf e_j=0$, which should imply vanishing Christoffel symbols (for $\nabla$ wrt the basis $\mathbf e_i$). But then again, they get Christoffel symbols in the expression for $\partial_i\mathbf V$, so my interpretation is that they are really dealing with two different covariant derivatives: the trivial one in the embedding space, which corresponds to the standard directional derivative, and has Christoffel symbols $\Gamma_{ij}^k$ used above, and this other $\nabla$, which is something different.
Which naturally spurs the question: can we always define such a covariant derivative which gives zero with respect to a given basis? If not, under what conditions can it be done?