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
If $f:(S,g)\to (S',g')$ is an isometry, then define
$\nabla_{X'}Y':=df\ \nabla_XY$
Show that this is LC-connection :
(1) Compatibility condition : First show that $$ X'(Y',Z')=X(Y,Z)$$
Proof : If $\frac{d}{dt}p(t)=X,\ p(0)=p$ then
$$ df_p X(df_p Y, df_p Z) =\frac{d}{dt} (df Y, df Z)_{f(p(t))} =
\frac{d}{dt} (Y,Z)_{p(t)}
$$ since $f$ is an isometry And $\frac{d}{dt} (Y,Z)_{p(t)}= X(Y,Z)$
So $$ (\nabla_{X'}Y',Z')+(Y',\nabla_{X'}Z')=f^\ast g'( \nabla_XY,Z)
+ f^\ast g' (Y,\nabla_XZ) = X(Y,Z) =X'(Y',Z') $$
(2) Symmetry condition : $$ \nabla_{X'}Y' -\nabla_{Y'}X'=df
(\nabla_XY-\nabla_YX)=df[X,Y]=[X',Y']$$
Best Answer
One way you can do this problem is with the Koszul Formula. Choose an arbitrary vector field $Z$, and use that formula to get
\begin{align} 2 \langle \nabla^2_X Y, Z \rangle_2 &= X \langle Y, Z \rangle_2 + Y \langle Z, X \rangle_2 - Z \langle X, Y \rangle_2\\ &+ \langle [X, Y], Z \rangle_2 - \langle [Y, Z], X \rangle_2 - \langle [X, Z] , Y \rangle_2. \tag{*}\label{*} \end{align}
Then \begin{align} X \langle Y, Z \rangle_2 = X(e^{2\rho}\langle Y, Z \rangle_1 ) &= e^{2\rho} X\langle Y, Z \rangle_1 + \langle Y, Z \rangle_1 X(e^{2\rho} )\\ &= e^{2\rho} X\langle Y, Z \rangle_1 + \langle Y, Z \rangle_1 e^{2\rho} \cdot 2 d\rho(X). \end{align}
We also have $$\langle [X, Y], Z \rangle_2 = e^{2\rho} \langle [X, Y], Z \rangle_1.$$
Applying identities of this form to each term of the right-hand side in \eqref{*}, we get \begin{align} 2 \langle \nabla^2_X Y, Z \rangle_2 = e^{2\rho} \cdot 2 \langle \nabla_X Y, Z \rangle_1 + 2 d \rho(X) e^{2\rho}\langle Y, Z \rangle_1 + 2 d \rho(Y) e^{2\rho}\langle Z, X \rangle_1 - 2 e^{2\rho} \langle X, Y \rangle_1 \langle \text{grad} \rho, Z \rangle_1 \end{align}
But $\langle \nabla^2_X Y, Z \rangle_2 = e^{2\rho} \langle \nabla^2_X Y, Z \rangle_1$, so we can cancel $2e^{2\rho}$ from both sides and get
\begin{align} \langle \nabla^2_X Y, Z \rangle_1 &= \langle \nabla_X Y, Z \rangle_1 + d \rho(X) \langle Y, Z \rangle_1 + d \rho(Y) \langle Z, X \rangle_1 - \langle X, Y \rangle_1 \langle \text{grad} \rho, Z \rangle_1 \\ &= \langle \left(\nabla_X Y + d \rho(X) Y + d\rho(Y)X - \langle X, Y \rangle_1 \text{grad} \rho \right), Z \rangle_1. \end{align}
Since this last equation holds for all $Z$, the two vector fields are equal.