Most of modern books on integration theory, when constructing the Lebesgue integral, do not introduce manifolds prior. The transformation for Lebesgue integrals can then be stated as follows:
Let $\Omega \subseteq \mathbb{R}^d$ be open and $\Phi \colon \Omega \rightarrow \Phi(\Omega)\subseteq\mathbb{R}^d$ a diffeomorphism. The function $f$ is integrable on $\Phi(\Omega)$ if and only if $x\mapsto f(\Phi(x))|\det(D\Phi(x))|$ is on $\Omega$. It then holds that
$\displaystyle \int_{\Phi(\Omega)}f(y)\,\mathrm{d}y=\int_{\Omega} f(\Phi(x))|\det(D\Phi(x))|\,\mathrm{d}x$
where $D\Phi(x)$ is the functional matrix.
When considering integrals over manifolds one could proceed like this (where some details are omitted for the purpose of transparency and only global charts are considered):
Let $\Psi\colon T\rightarrow V\subseteq M, T\subseteq\mathbb{R}^k$ be a global chart.
The integral of $f$ over $M$ is then defined as
$\displaystyle \int_M f(x)\,\mathrm{d}S(x)=\int_T f(\Psi(t))\sqrt{g(t)}\,\mathrm{d}t$
where $g$ is the gramian determinant of the chart $\Psi$.
I am looking for an analogue of the first integral transformation for manifolds. Has this anything to do with a change of charts? Also, if I remember correctly, in the theory of differentialforms this analogue is the pullback of a form, is this correct? Do you have to invoke parametrisations for a pullback, too?
Given a parametric $k$-dimensional manifold $\Psi \colon V \rightarrow \mathbb{R}^n$ where $V \subseteq \mathbb{R}^k$ is an open subset and $\Psi$ satisfies the usual conditions (one-to-one, smooth, full-rank), the integral over $M = \Psi(V)$ of a function $f \colon M \rightarrow \mathbb{R}$ is defined by
$$ \int_{M} f \, dS := \int_V f(\Psi(y)) \, \sqrt{\det (D\Psi^T(y)\cdot D\Psi(y))} \, dy. $$
To make sure this definition makes sense, we need to verify that it is actually independent of the specific parametrization (chart) of $M$. If $\Phi \colon V' \rightarrow \mathbb{R}^n$ is another parametrization of $M$ then $g := \Psi^{-1} \circ \Phi \colon V' \rightarrow V$ is a diffeomorphism between open sets $V,V' \subseteq \mathbb{R}^k$ and the regular change of variables formula implies that
$$ \int_V f(\Psi(y)) \, \sqrt{\det \left( D\Psi^T(y) \cdot D\Psi(y) \right)} \, dy = \int_{V'} f(\Psi(g(x))) \sqrt{\det \left( D\Psi^T(g(x)) \cdot D\Psi(g(x)) \right)} \left| \det Dg(x) \right| dx $$
Note that
$$ \Psi(g(x)) = \Psi(\Psi^{-1}(\Phi(x))) = \Phi(x), \\ f(\Psi(g(x)) = f(\Phi(x)), \\ D\Phi(x) = D\Psi(g(x)) \cdot Dg(x), \\ \sqrt{\det \left( D\Psi^T(g(x)) \cdot D\Psi(g(x)) \right)} \left| \det Dg(x) \right| = \\ \sqrt{ \det(Dg^T(x)) \det \left( D\Psi^T(g(x)) \cdot D\Psi(g(x)) \right) \det(Dg(x)) } = \\ \sqrt{\det \left( Dg^T(x) \cdot D\Psi^T(g(x)) \cdot D\Psi(g(x)) \cdot Dg(x) \right)} = \\ \sqrt{ \det \left( D\Phi^T(x) \cdot D\Phi(x) \right) } $$
and so we also get
$$ \int_V f(\Psi(y)) \, \sqrt{\det \left( D\Psi^T(y) \cdot D\Psi(y) \right)} \, dy = \int_{V'} f(\Phi(x)) \, \sqrt{ \det \left( D\Phi^T(x) \cdot D\Phi(x) \right) } \, dx. $$
In other words, the invariance of the integral where changing charts follows from the regular change of variables rule. Note that if $k = n = d$, the independence of the definition on the chart is precisely the statement of the change of variables rule.