If $ \phi : {\bf R}^n \rightarrow M$ is coordinate, then $$
\int_{U:=\phi ({\bf R}^n) } \Phi (p)\ d {\rm Vol}_p := \int_{{\bf
R}^n} \Phi (\phi(x)) \sqrt{{\rm det}\ g_{ij}} dx^1\cdots dx^n
$$
where $$ g_{ij}:=g(d\phi e_i,d\phi e_j ) $$
If $ f : x\in {\bf R}^n\rightarrow y\in {\bf R}^n $ is a diffeomorphism, we
have another coordinate $$ \phi\circ f^{-1} : {\bf R}^n\rightarrow
M $$
That is by above definition we have $$ \int_{U} \Phi (p)\ d {\rm
Vol}_p = \int_{{\bf R}^n} \Phi (\phi\circ f^{-1} (y) ) \sqrt{{\rm
det}\ g_{\alpha\beta}} dy^1\cdots dy^n
$$ where $$ g_{\alpha\beta}:= g(d(\phi\circ f^{-1}) e_\alpha,d(\phi\circ f^{-1} ) e_\beta
) $$
Here $$ f^\ast ( \Phi (\phi\circ f^{-1} (y) ) \sqrt{{\rm det}\
g_{\alpha\beta}} dy^1\cdots dy^n ) $$ $$= \Phi (\phi (x)) \sqrt{
{\rm det}\ (f^{-1})^k_\alpha (f^{-1})^l_\beta g_{kl} }\ {\rm det}
f^\alpha_k\ dx^1\cdots dx^n $$ $$= \Phi (\phi (x)) \sqrt{ {\rm det}\
g_{kl} }\ dx^1\cdots dx^n$$
That is since two integrals are equal (cf. change of variables), the above definition is well
defined.
The basic idea is the following.
For each point $p\in N$, the fiber $f^{-1}(p)$ is a set of $n$ points in $M$. Each such point has a neighbourhood such that the restriction of $f$ to it is an isometry. Now take the intersection of the images of these sets, which will be some open neighbourhood $p \in U \subset N$. By its definition, the restriction of $f$ to each connected component of $f^{-1}(U)$ will be an isometry. I claim you can prove the following (let $\nu$ denote the volume form on $N$ corresponding to the metric):
Lemma
$$Vol(f^{-1}(U)) = \int_{f^{-1}(U)}f^*(\nu) = n\cdot \int_{U}\nu = n\cdot Vol(U)$$
In fact, we have the following slightly stronger statement:
Lemma For any function $\psi$ and a volume form $\nu$,
$$\int_{f^{-1}(U)}f^*(\psi\nu) = n\cdot \int_{U}\psi\nu $$
This lemma is a local version of the statement you want. We used the local isometry property here. We would now like to make it global. How do you typically achieve such a thing in differential geometry? One common strategy is to use a partition of unity.
Take a cover of $N$ by open sets $U_\alpha$, such that for each $U_\alpha$, the restriction of $f$ to each connected component of $f^{-1}(U_\alpha)$ is an isometry. (I claim you can do this, but you should check carefully that it is true).
Take a partition of unity $\{ \psi\}$ on $N$ that is subordinate to your chosen open cover (I omit indices). The volume form for our Riemannian metric, $\nu$, is equal to
$$ \nu = \sum \psi \cdot \nu.$$
(where the sum is over functions in our partition). Now
$$ Vol(M) = \int_M f^*\nu = \int_M f^*\left(\sum \psi \cdot \nu\right) = \sum\int_M f^* \left(\psi \cdot \nu\right) = \sum\int_{f^{-1}(U_\alpha)} f^* (\psi \cdot \nu)$$
(if your volume is infinite there might be some convergence issues around commuting the sum and integral, but this case can probably be handled separately). By our lemma,
$$ \int_{f^{-1}(U_\alpha)} f^* \psi \cdot \nu = n \cdot \int_{U_\alpha} \psi \cdot \nu $$
Putting this together with the fact that $\psi$ is a partition of unity, we continue from above:
$$\sum\int_{f^{-1}(U_\alpha)} f^* (\psi \cdot \nu) = n\sum \int_{U_\alpha} \psi \cdot \nu = n \int_{M} \sum\psi \cdot \nu =n \int_{M} \nu = n Vol(N)$$
Best Answer
I'm not sure I understand your objection fully. So, you're starting with two (pseudo)-Riemannian metrics $g_1,g_2$ on the same manifold $M$, and these induce measures $\lambda_{g_1},\lambda_{g_2}$ on $M$. Note that the very definition of these measures is via charts, so let's just record this explicitly:
This definition/theorem is exactly how we characterize the measure $\lambda_g$, so there's no way to avoid charts completely (after all why would you; the above definition is literally how one calculates integrals over many of the standard manifolds in practice). Now, as long as we can show that the two measures are mutually absolutely continuous, then the Radon-Nikodym theorem tells us there exists a measurable, non-negative function $\rho:M\to [0,\infty]$ such that $d\lambda_{g_2}=\rho\,d\lambda_{g_1}$.
In other words, all you have to do is show that for any Lebesgue-measurable subset $A\subset M$, if $\lambda_{g_1}(A)=0$ then $\lambda_{g_2}(A)=0$. Since the manifold can be covered by countably many charts $(U_i,x_i)$, it suffices to prove each $\lambda_{g_2}(A\cap U_i)=0$. In other words, we may as well assume at the outset that the set $A$ is contained inside a single coordinate chart $(U,x)$. Now, we have \begin{align} 0&= \lambda_{g_1}(A)=\int_{x[A]}\sqrt{|\det g_{1}|}\,d\lambda_n \end{align} First equality is by assumption, the second is by definition of $\lambda_{g_1}$. Now, $g_1$ being a pseudo-Riemannian metric implies $\sqrt{|\det g_1|}$ is a positive function on $x[U]$, so if the Lebesgue integral of a positive function vanishes, then the set, $x[A]$, being integrated over is a $\lambda_n$-null set. Thus, \begin{align} \lambda_{g_2}(A)&:=\int_{x[A]}\sqrt{|\det g_2|}\,d\lambda_n=0, \end{align} because as we showed above, the set, $x[A]$, being integrated over is a $\lambda_n$-null set.
This completes the proof. Just to reiterate, what we've shown is that the two measures are mutually (since you can interchange $g_1,g_2$) absolutely continuous, so by the Radon-Nikodym theorem such a $\rho$ exists. This is already a globally defined-function; there are no more well-definition checks to be made. What your argument shows is that this $\rho$ is given by a quotient of determinants, so by the explicit formula you have, you can deduce that $\rho$ is a smooth function $M\to (0,\infty)$; this is more than we can say by just appealing to the Radon-Nikodym theorem.
Of course, another approach to all of this is to simply check that the quotient you have doesn't depend on the choice of charts (this is because if you change between $x$ and $y$ coordinates the numerator and denominator both pick up a factor of $|\det D(y\circ x^{-1})|$, which cancels out). As a result, these locally defined functions can be patched together to form a globally-defined smooth function $\rho: M\to (0,\infty)$, and this is done without ever invoking the Radon-Nikodym theorem.