Ask for an explicit proof of linear lemma in Riemannian geometry

Lemma Let $f:(M,g)\rightarrow(\bar{M},\bar{g})$ be an isometry between two Riemannian manifolds, then $df(\nabla_X Y)=\bar{\nabla}_{df(X)} df(Y)$ where $\nabla,\bar{\nabla}$ are Riemannian connections of $M,\bar{M}$ respectively.

This is called by my tutor the 'linear lemma'. He said that this could be easily perceived by linearity. But when I tried to prove it using Koszul formula for Riemannian connection, I still find it not clear somewhere in my calculation.

Here is my calculation: $2\bar{g}(df(\nabla_X Y),df(Z))$ $=2g(\nabla_X Y,Z)$(isometry definition) $=Xg(Y,Z)+Yg(Z,X)-Zg(X,Y)-g(X,[Y,Z])+g(Y,[Z,X])+g(Z,[X,Y])$(Koszul formula) $=X(\bar{g}(df(Y),df(Z))\circ f)-\ldots-\bar{g}(df(X),df[Y,Z])\circ f$(isometry definition) $=(df(X)\bar{g}(df(Y),df(Z)))\circ f-\ldots-\bar{g}(df(X),[df(Y),df(Z)])\circ f$

$X(\bar{g}(df(Y),df(Z)))-\bar{g}(df(X),df([Y,Z]))=df(X\bar{g}(df(Y),df(Z)))-\bar{g}(df(X),[df(Y),df(Z)])$(A direct calculation)

(Here is my problem, I do not know if $df[X,Y]=[df(X),df(Y)]$ is correct, just seemingly correct by linearity) Assuming this true:

$=2\bar{g}(\nabla_{df(X)} df(Y),df(Z))\circ f$(Koszul formula)

Where can I find an explicit proof of this lemma? I have not found in do Carmo's book and Petersen's book.(It is an exercise in the latter book)

Thanks for your time!

$df[X,Y] = [df(X), df(Y)]$ is always true when $f$ is a diffeomorphism between 2 smooth manifolds. The principle is that $df : \Gamma(M) \to \Gamma(\bar{M})$ makes $X \in \Gamma(M)$ and $\bar{X} = df(X) \in \Gamma(\bar{M})$ $f$-related by definition for any $X$.
$$df \circ X = \bar{X} \circ f$$