The Lie derivative $L_X T$ of any tensor $T$ along any vector field $X$ is defined directly and only from the underlying manifold structure. If, moreover, the manifold has a symmetric connection, (symmetric means zero torsion), then it is possible to express $\mathcal{L}_X T$ using this connection. Of course, the result is the same no matter which connection is used, (as long as it is symmetric).
Please note, accordingly, there is no meaning in writing $L_X$ and $\bar{L}_X$. This is wrong. The Lie derivative comes first, and then the expression in terms of a connection, not the other way around.
The Lie derivative has two (equivalent) definitions. A dynamical one and an algebraic one. It is very important to appreciate that these reflect two equally useful intuitions. I will not get into this here, (feel free to ask), but I will consider expressing the Lie derivative in terms of a symmetric connection.
If $\nabla$ is a symmetric connection, then for any vector fields $X,Y$
$$
L_XY = [X,Y] = \nabla_XY-\nabla_YX
$$
where the first equality is by definition, but the second equality means $\nabla$ is symmetric.
For $(0,s)$-tensor $T$, (think of the metric if you would like), $L_XT$ is also a $(0,s)$-tensor, (one says that $L_X$ is type preserving), and by definition
$$
L_XT(Y_1,\ldots,Y_s) = X(T(Y_1,\ldots,Y_s)) - \sum_i T(\ldots,[X,Y_i],\ldots)
$$
The first term is
$$
X(T(Y_1,\ldots,Y_s)) = (\nabla_X T)(Y_1,\ldots,Y_s) + \sum_i T(\ldots,\nabla_XY_i,\ldots)
$$
By replacing in the definition of $L_XT$ and using the fact that $\nabla$ is symmetric, this yields the expression
$$
L_XT(Y_1,\ldots,Y_s) = \nabla_X T(Y_1,\ldots,Y_s) + \sum_i T(\ldots,\nabla_{Y_i}X,\ldots)
$$
Please note this is not the definition of $L_XT$ but only a formula, using the connection $\nabla$. In other words, $L_XT$ does not depend on $\nabla$ but the righ-hand side does.
If $T$ is parallel, (this means $\nabla_X T = 0$ for any $X$),
$$
L_XT(Y_1,\ldots,Y_s) = \sum_i T(\ldots,\nabla_{Y_i}X,\ldots)
$$
You may apply this last formula to the Lie derivative of the Riemannian metric $T=g$, and use the Levi-Civita connection as $\nabla$, to get the ''elasticity tensor'' $L_Xg$ and understand the definition of Killing vector fields, etc.
-- Salem
By an equivalent definition, inner product (contraction) of two $(0,2)$ tensors $A_{ij}$ and $B_{ij}$ is as follows:
$$\langle A,B\rangle= A_{i}{\ }^{j}B_{j}{\ }^{i}.$$ and it is also equal to $A^{i}{\ }_{j}B^{j}{\ }_{i}$. One can show that
$$\langle A,B\rangle= A(e_i,e_j)B(e_i,e_j)$$
for arbitrary orthonormal frame $\{e_1,\dots,e_n\}$.
By this we have
$$\langle df\otimes df,B\rangle=(df\otimes d f)(e_i,e_j)A(e_i,e_j)=df(e_i)df(e_j)A(e_i,e_j)=(e_if)(e_jf)A(e_i,e_j)=A((e_if)e_i,(e_jf)e_j)=A(\nabla f,\nabla f).$$
Note that in general by the definition of $df$ $$df(X)=X.f=\nabla_Xf$$
and $(e_if)e_i=\nabla f=\mathsf{grad} f=df^\sharp$. Now let $A=\mathcal{L}_Xh$.
Now can you prove @Jesse Madnick claim in comment? (Proof is same as above.) $$\langle \alpha \otimes \alpha, \beta \rangle = \beta(\alpha^\sharp, \alpha^\sharp)$$
Best Answer
This formula is a concise and expressive version of Koszul formula. This is just the matter of regrouping the terms.
It shows that the Levi-Civita covariant derivative is given with a formula, which employs only the Lie derivative, the exterior derivative, and the given Riemannian metric.
I find this formula very illuminating, because the Lie derivative and the exterior derivative are always present on a smooth manifold (do not require any choice), and the only choice is made when a Riemannian metric is fixed.
This formula also exhibits the important properties of the covariant derivative: it is linear in $Y$ and non-linear in $X$.
Furthermore, in reveals that the Levi-Civita covariant derivative depends on the Lie derivative of the metric, which is the source of non-linearity on the slot $X$. This observation may also lead to other interesting insights.