$g^*(\mbox{d}x) = \mbox{d}x \circ g^{-1}$ for $g$ an isometry Suppose $(M,\gamma)$ is a Riemannian manifold, $g$ an isometry, $\mbox{d}x$ the Riemannian volume form on $(M, \gamma)$. I can't really understand these formulae, sometimes found in literature. First,
\begin{equation}\tag{1}
g^*(\mbox{d}x) = \mbox{d}x\circ g^{-1}.
\end{equation}
Next, let $f : M \to N$ be a smooth function (here, $g$ can be seen as an element of a group $G$ acting by isometries in some specified way on both $M$ and $N$); sometimes I find
\begin{equation}\tag{2}
\vert{\mbox{d}(g \circ f)}\vert^2 = \vert \mbox{d}f\vert^2 \circ g^{-1}.
\end{equation} 
I can't catch the precise meaning of the rhs. I know what the operations involved are (pullback, exterior derivative, commutation property of pullback and exterior derivative, differential, chain rule...); my problem concern what one precisely means when writing a thing such as $g^*(\mbox{d}x) = \mbox{d}x \circ g^{-1}$ instead of 
\begin{equation}
g^*(\mbox{d}x) = \sqrt{|\gamma(g(x))|}J_g(x)\mbox{d}g^1(x) \wedge\dots \wedge \mbox{d}g^m(x) = \mbox{d}(g(x)).
\end{equation}
Here, $m = \dim M$, $J_g(x)$ the jacobian of the coordinate transformation $g$ evalueted at $x$ while $\sqrt{|g^*(\gamma(x))|} = \sqrt{|\gamma(g(x))|}$ because $g$ is an isometry.
I'm not really into differential and Riemannian geometry, so I'm probably missing something entirely elementary.
 A: I suppose $(1)$ refers to the usual definition of pull-back for tensors (and hence differential forms):
if $\omega$ is a $n-$form on $N$ and $F\colon M\to N$ is $C^{\infty}$, then
$$
(F^*\omega)_p(v_1,\dots,v_n):=\omega_{F(p)}(dF_p(v_1),\dots,dF_p(v_n)).
$$
which actually is what you have already written in the last part. So, actually it is
$$
F^*\omega=\omega\circ F,
$$
yet one needs to evaluate $\omega$ at vectors in $T_N$, and hence one uses the differential of $F$.
EDIT:
$(2)$ means that
$$
||d(g\circ f)||_x^2=||df||^2_{g^{-1}(x)},
$$
where $||\cdot||$ is the norm induced by the bundle metric on $T^*_M$ (see Riemannian Geometry and Geometric Analysis, Jost Theorem 2.1.4) given in coordinates by
$$
||\omega||:=|g^{ij}\omega_i\omega_j|^{1/2},
$$
where $\omega=\omega_idx^i$ (Einstein summation is being used). Note that $(2)$ is equivalent to
$$
||\nabla(g\circ f)||_x^2=||\nabla f||^2_{g^{-1}(x)}.
$$
Since $g$ is an isometry, this looks fairly reasonable. I suppose that some calculations (repeated use of the chain rule) would show that it is true.
