Let $S^2$ denote the unit 2-sphere, $M$ be a 2-dimensional oriented embedded $C^1$-submanifold of $\mathbb R^3$ with $$d_M(x,\omega):=\inf\left\{t>0:x+t\omega\in M\right\}<\infty\;\;\;\text{for all }x\in M\text{ and }\omega\in S^2\tag1,$$ $\nu_M$ denote the outer normal field on $M$, $\sigma_M$ denote the surface measure on $\mathcal B(M)$, $$S^2(x):=\left\{y\in\mathbb R^3:|x-y|=1\right\}=x+S^2\;\;\;\text{for }x\in\mathbb R^3$$ and $$\omega_{x\to y}:=\frac{y-x}{|y-x|}\in S^2\;\;\;\text{for }x,y\in\mathbb R^3\text{ with }x\ne y.$$
Let $x\in M$. Can we show that the pushforward measure of $\sigma_{S^2}$ under $$\varphi:S^2\to M\;,\;\;\;\omega\mapsto x+d_M(x,\omega)\omega$$ has a density with respect to $\sigma_M$?
The desired the claim should be true. I guess we need to use $$\sigma_{S^2}(B)=\int\sigma_M({\rm d}y)1_B\left(\omega_{x\to y}\right)\frac{\left|\langle\nu_M(y),\omega_{x\to y}\rangle\right|}{|x-y|^2}\;\;\;\text{for all }B\in\mathcal B(S^2)\tag2$$ (I've asked for that separetely: Can we prove this relation between the solid angle measure and the surface measure of a smooth manifold?). Clearly, if $x\in\mathbb R^3$, $\varepsilon>0$ and $f:\varepsilon S^2(x)\to\mathbb R$ with $f\ge0$ or $\int|f|\:{\rm d}\sigma_{\varepsilon S^2(x)}<\infty$, then $$\int f\:{\rm d}\sigma_{\varepsilon S^2(x)}=\varepsilon^2\int\sigma_{S^2}({\rm d}\omega)f(\varepsilon(x+\omega))\tag3.$$ However, if $B\in\mathcal B(M)$, we cannot immediately apply this to $$\left(\sigma_{S^2}\circ\varphi^{-1}\right)(B)=\int\sigma_{S^2}({\rm d}\omega)1_B\left(d_M(x,\omega)\left(\frac x{d_M(x,\omega)}+\omega\right)\right)\tag4.$$
