If $g_i : {\bf R}^n\rightarrow M $ is orientable chart for $M$, then for any two, $$ {\rm det}\ d(g_2^{-1}\circ g_1 ) > 0 $$ since $M$ is orientable
So $h_i:=f\circ g_i : {\bf R}^n\rightarrow N $ is chart so that $$
{\rm det}\ d( h_2^{-1}\circ h_1) = {\rm det}\ d(g_2^{-1}\circ f^{-1} \circ f \circ g_1)= {\rm det}\ d(g_2^{-1}\circ g_1) >0 $$
Yeah, and you can make it work even if $f$ is not a local diffeomorphism, only a proper map. The relevant notion is that of the degree of a smooth proper map between oriented manifolds of the same dimension.
Assume for simplicity that $M,N$ are closed (compact, without boundary), non-empty and connected and let $f \colon M \rightarrow N$ be an arbitrary smooth map. Since $M,N$ are oriented, we can a generator $[\omega_1] \in H^{\text{top}}(M)$ such that $\omega_1$ is consistent with the orientation on $M$ and $\int_M \omega_1 = 1$. Choose $[\omega_2]$ similarly for $N$. The map $f$ induces a map $f^{*} \colon H^{\text{top}}(N) \rightarrow H^{\text{top}}(M)$ on cohomology and since the top cohomology groups are one-dimensional, we must have $f^{*}([\omega_2]) = c [\omega_1]$ for some $c \in \mathbb{R}$. This $c = \deg(f)$ is called the degree of $f$ and is a priori a real number. However, it can be shown that $c$ is in fact an integer which can be computed by counting the number of preimages of a regular value $p \in N$ with appropriate signs which take the orientations of $M,N$ into consideration.
Knowing that, given $\omega \in \Omega^{\text{top}}(N)$, write $[\omega] = c[\omega_2]$ for some $c \in \mathbb{R}$. Then
$$ [f^{*}(\omega)] = f^{*}([\omega]) = f^{*}(c[\omega_2]) = c \deg(f) [\omega_1] $$
so
$$ \int_M f^{*}(\omega) = \deg(f) c \int_N \omega_1 = \deg(f) c = \deg(f) c\int_N \omega_2 = \deg(f) \int_N \omega. $$
In particular, if $f$ is a diffeomorphism, $\deg(f) = \pm 1$ so this generalizes your starting point. If $f$ is a local diffeomorphism then $f \colon M \rightarrow N$ is a covering map and $\deg(f)$ will be the number of points in an arbirary fiber, counted with appropriate signs. For much more details and proofs, see the book "Differential Forms in Algebraic Topology".
Best Answer
I think I have a elementary way to solve, making calculus
$x_\beta^{-1}\circ x_\alpha = (\phi^{-1}\circ y_\beta)^{-1}\circ(\phi^{-1}\circ y_\alpha) = ((\phi^{-1}\circ y_\alpha)^{-1}\circ(\phi^{-1}\circ y_\beta))^{-1} = (y_\alpha^{-1}\circ y_\beta)^{-1} = y_\beta^{-1}\circ y_\alpha$
So, getting de determinant of the differential of change of coordinates in M1 is the same as in M2, so is just to make sure that all is well defined.
If I am right then a question come to my mind, the other implication could be true?, is M1 is orientable then M2 is orientable?