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:
Let $(M,g)$ be an $n$-dimensional pseudo-Riemannian manifold, and $\mathcal{L}_M$ the Lebesgue $\sigma$-algebra on $M$. Then, there is a unique positive-measure $\lambda_g$ defined on the $\sigma$-algebra $\mathcal{L}_M$ such that for every $A\in\mathcal{L}_M$ and every chart $(U,x)$, if $A\subset U$ then $\lambda_g(A)=\int_{x[A]}\sqrt{|\det g_{i,j}|}\, d\lambda_n$ ($\lambda_n$ refers to Lebesgue measure on $\Bbb{R}^n$).
In fact, once can easily show $\lambda_g$ is complete, $\sigma$-finite and yields a positive number on any non-empty open set.
Remarks:
- Here $M$ is assumed second-countable. Orientability is unimportant. Partitions of unity are also not necessary
- We say $A\in\mathcal{L}_M$ if and only if for every chart $(U,x)$ of the manifold, $x[A\cap U]$ is a Lebesgue-measurable subset of $\Bbb{R}^n$. It is easily verified that $\mathcal{L}_M$ is indeed a $\sigma$-algebra (which contains the Borel $\sigma$-algebra of $M$).
- This uniquely defined measure is called the Riemann-Lebesgue volume measure on $M$ induced by $g$.
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.
