stevendaryl said:
I guess
@PeterDonis would say that this ambiguity is resolved by denying that ##\nabla_\mu## is an operator
Not necessarily, no. With your convention, where ##\nabla_\mu## means ##\nabla_{e_\mu}##, the directional derivative operator along ##e_\mu##, it is obviously an operator. With the convention I'm used to, ##\nabla_\mu##, used in isolation, is just a way of referring to the covariant derivative operator ##\nabla## itself--Wald would write it as ##\nabla_a##. But in either case it's an operator. The ambiguity in the convention I'm used to is, as you say, that sometimes (usually in expressions where it's combined with other things), ##\nabla_\mu## can mean the ##\mu## component of some tensor obtained by applying the ##\nabla## operator to something; that is indeed not an operator. (As we have seen, it ends up being the same as the directional derivative in the ##e_\mu## direction of the thing the ##\nabla## is operating on. But that still doesn't resolve all ambiguities; see below.)
The only real way to resolve ambiguity is to, well, resolve ambiguity, by adding more notation until the expression is unambiguous.
For example, in Wald's abstract index notation, the various objects you have given would look like this:
Directional derivative of ##V##:
$$
\nabla_\mu V = \left[ \left( e_\mu \right)^a \nabla_a \right] V^b
$$
(Note the brackets enclosing the contraction that denotes the directional derivative, to make it unambiguous that it denotes a single operator.)
Extracting the ##\nu## component:
$$
\left( \nabla_\mu V \right)^\nu = \left[ \left( e_\mu \right)^a \nabla_a V^b \right] \left( e^\nu \right)_b
$$
(Here I don't have a third bracket type to use, so I'm relying on the first expression above to make it clear what the directional derivative operator is, and using the brackets to make clear that the operator is only operating on ##V##; the operator produces a vector, and we contract that vector with ##e^\nu## to extract the component.)
Directional derivative of the ##\nu## component of ##V##:
$$
\nabla_\mu V^\nu = \left[ \left( e_\mu \right)^a \nabla_a \right] \left[ V^b \left( e^\nu \right)_b \right]
$$
Expanding out the above (since now the directional derivative is operating on both ##V## and ##e_\nu##, as the brackets in the above expression make clear):
$$
\nabla_\mu V^\nu = \left[ \left( e_\mu \right)^a \nabla_a V^b \right] \left( e^\nu \right)_b + V^b \left[ \left( e_\mu \right)^a \nabla_a \left( e^\nu \right)_b \right]
$$
Notice that this does
not give the same result as above.
Covariant derivative of ##V##:
$$
\nabla V = \nabla_a V^b
$$
Extracting the ##\mu##, ##\nu## component:
$$
\left( \nabla V \right)_\mu{}^\nu = \left( \nabla_a V^b \right) \left( e_\mu \right)^a \left( e^\nu \right)_b
$$
As we have seen in previous posts, since covariant differentiation commutes with contraction, this is equal to the ##\nu## component of the directional derivative of ##V## in the ##e_\mu## direction, but it is
not equal to the directional derivative in the ##\mu## direction of the ##\nu## component of ##V##.