Confusion about Lie derivative on metric

This is the first time I face this problem, I think that the answer is positive provided the connection is torsion free as, in fact, happens for the Levi-Civita connection associated with a metric. Here is my proof.

Let us indicate the Lie derivative with respect to the vector field $Z$ by ${\cal L}_Z$. The action on tensor fields is completely determined by the following requirements.

[0] ${\cal L}_Z(X) = [Z,X]$.

[1] It trivially acts on scalar fields: ${\cal L}_Z f = Z(f)$.

[2] It commutes with contractions: ${\cal L}_Z(\langle X, \omega \rangle) = \langle {\cal L}_Z(X), \omega \rangle + \langle X, {\cal L}_Z(\omega) \rangle$.

[3] It acts as a derivative with respect to the tensor product: ${\cal L}_Z (T\otimes U) = ({\cal L}_Z(T) )\otimes U + T \otimes {\cal L}_Z(U)$.

The requirements [0] and [2] completely define the actions on covariant vector fields in terms of the action of vector fields on scalar fields (components of covariant fields): $${\cal L}_Z(\omega) = \langle \partial_{x^a}, {\cal L}_Z(\omega) \rangle dx^a = Z(\langle \partial_{x^a}, \omega\rangle ) dx^a - \langle Z(\partial_{x^a}), \omega\rangle dx^a = Z(\omega_a) dx^a - \langle [Z, \partial_{x^a}], \omega \rangle dx^a\:,$$ i.e., $${\cal L}_Z(\omega)_a = Z(\omega_a) - \langle [Z, \partial_{x^a}], dx^b \rangle \omega_b\:. \tag{1}$$

From the definition of torsion we have that, if $\nabla$ is a torsion-free connection, then $$[Z,X] = \nabla_Z X - \nabla_X Z$$
since the torsion is nothing but the difference of the two sides. From [0] we conclude that $${\cal L}_Z(X) = \nabla_Z X - \nabla_X Z\:, \tag{2}$$ or, in components: $$({\cal L}_Z(X))^a = (Z^b\nabla_b X)^a - (\nabla_b Z)^aX^b\:,$$ and this identity just says that, for torsion-free connctions, we can replace derivatives for covariant ones when computing the Lie derivative of contravariant vector fields.

Let us pass to covariant vector fields. From (1) and (2) we have that $${\cal L}_Z(\omega)_a = \nabla_Z(\omega_a) - \langle \nabla_Z( \partial_{x^a}), dx^b \rangle \omega_b + \langle \nabla_{\partial_{x^a}}(Z), dx^b \rangle \omega_b\:,$$ that is $${\cal L}_Z(\omega)_a = (Z^b\nabla_b\omega)_a + \langle \nabla_{a}(Z), dx^b \rangle \omega_b\:,$$ namely $${\cal L}_Z(\omega)_a = (Z^b\nabla_b\omega)_a + (\nabla_{a}Z)^b\omega_b\:,$$ and this identity just says that, for torsion-free connctions, we can replace derivatives for covariant ones when computing the Lie derivative of covariant vector fields.

Since generic tensor fields can be constructed as a linear combination of tensor products of contravariant and covariant vector fields, from [3] the found results can easily be generalized to all types of tensor fields. In computing the Lie derivative of a tensor field we can always replace the standard derivative for the covariant one at every occurrence provided the connection is torsion free.

For any vector field $X$, $\nabla_X$ is a derivation of the tensor algebra and thus can be decomposed as $$ \nabla_X = \mathcal L_Y + S $$ for some vector field $Y$ and an endomorphism $S$ of the tangent space (see eg Kobayashi/Nomizu, proposition 3.3).

For a function $f$, we have $$ \nabla_X f = \mathcal L_X f $$ and thus $Y=X$.

Given an arbitrary vector field $Z$, by definition of torsion $$ \nabla_X Z = \mathcal L_X Z + \nabla_Z X + T(X,Z) $$ In case of torsion-free connections, this yields $$ S: Z \mapsto \nabla_Z X $$ ie $$ \mathcal L_X = \nabla_X - \nabla_{(\cdot)} X $$ As $\partial$ is just a local connection on the coordinate patch, this actually yields both coordinate expressions mentioned in the question.

Note that in case of covariant tensors, the positive sign is due to the fact that covectors need to transform with the negative transpose of the endomorphism (see eg Kobayashi/Nomizu, proposition 2.13).