Let $d\in\mathbb N$, $k\in\{1,\ldots,d\}$ and $\Omega$ be a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$. Assume, for simplicity, that $\Omega$ is described by a single chart, i.e. there is a $C^1$-diffeomorphism from $\Omega$ onto an open subset $U$ of $\mathbb R^k$.
Remember the definition of the surface measure $$\sigma_\Omega:=\phi^{-1}_\ast\left(\det\left|{\rm D}\phi^{-1}\right|\left.\lambda^{\otimes k}\right|_U\right),$$ where $\lambda$ denotes the Lebesgue measure on $\mathcal B(\mathbb R)$ and $|A|:=\sqrt{A^\ast A}$ for any matrix $A$.
Now let $T$ be a $C^1$-diffeomorphism from $\mathbb R^d$ onto $\mathbb R^d$. It's easy to see that $\Omega':=T(\Omega)$ is again a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ and $\phi':=\phi\circ\left.T^{-1}\right|_{\Omega'}$ is a chart for $\Omega'$.
If $f:\Omega'\to\mathbb R$ is $\sigma_{\Omega'}$-integrable, are we able to show that $$\int f\:{\rm d}\sigma_{\Omega'}=\int_U\det\left|{\rm D}{\phi'}^{-1}\right|\left(f\circ{\phi'}^{-1}\right){\rm d}\lambda^{\otimes k}=\int_U\det\left|{\rm D}\phi^{-1}\right|\left(g\circ\phi^{-1}\right){\rm d}\lambda^{\otimes k}\tag1$$ for some $\sigma_\Omega$-integrable $g:\Omega\to\mathbb R$?
Note that ${\phi'}^{-1}=T\circ\phi^{-1}$ and $${\rm D}{\phi'}^{-1}(u)={\rm D}T\left(\phi^{-1}(u)\right){\rm D}\phi^{-1}(u)\;\;\;\text{for all }u\in U.\tag2$$
Clearly, it's tempting to choose $g:=\left.\left|\det{\rm D}T\right|(f\circ T)\right|_\Omega$, but it seems like this choice doesn't yield the claim, when $k<d$.
If $k=d$, it's easy to solve this issue, since we are then able to write $$\det\left|{\rm D}{\phi'}^{-1}(u)\right|=\left|\det{\rm D}{\phi'}^{-1}(u)\right|=\left|\det{\rm D}T\left(\phi^{-1}(u)\right)\right|\left|\det{\rm D}\phi^{-1}(u)\right|\tag3$$ for all $u\in U$ and hence $$\int f\:{\rm d}\sigma_{\Omega'}=\int\left|\det{\rm D}T\right|(f\circ T){\rm d}\sigma_\Omega\tag4.$$
So, are we only able to find a transformation formula in the case $k=d$?