Let $M$ be a 2-dimensional embedded $C^1$-submanifold of $\mathbb R^3$ with a global chart$^1$ $(U,\phi)$. If $u\in U$ and $x=\phi^{-1}(u)$, let $\nu_M(x)$ denote the unique unit normal vector of $M$ with $$\det\left({\rm D}\phi(u),\nu_M(x)\right)>0\tag1.$$ Moreover, let $\sigma_M$ denote the surface measure$^2$ on the Borel $\sigma$-algebra $\mathcal B(M)$ and $$\pi:\mathbb R^3\setminus\{0\}\to S^2\;,\;\;\;x\mapsto\frac x{|x|}$$ denote the projection onto the unit 2-sphere $S^2$.

If $0\not\in B\in\mathcal B(M)$, are we able to express $\sigma_{S^2}(\pi(B))$ as an integral with respect to $\sigma_M$?

For clarity of exposition, let $$\omega_{x\to y}:=\pi(y-x)\;\;\;\text{for }x,y\in\mathbb R^3\text{ with }x\ne y.$$ There are plenty of (mathematically vaguely) references$^3$ claiming that $$\sigma_{S^2}({\rm d}\omega_{x\to y})=\sigma_M({\rm d}y)\frac{\left|\langle\nu_M(y),\omega_{x\to y})\rangle\right|}{|x-y|^2}\tag2,$$ which is reasonable from geometric inspection. However, I struggle to state and prove this in a measure-theoretic rigorous way.

**Idea 1**: Noting that $\pi=\nabla\rho$, where $\rho(x):=|x|$ for $x\in\mathbb R^3$, $(2)$ may be related to the divergence theorem. To be precise, if $K\subseteq M\setminus\{0\}$ is compact and has a $C^1$-boundary$^2$, $$\langle\nu_{\partial K},\pi\rangle=\langle\nu_{\partial K},\nabla\rho\rangle=:\frac{\partial\rho}{\partial\nu_{\partial K}}.\tag3$$ However, I'm unsure how the "outer" normal field$^5$ $\nu_{\partial K}$ and $\nu_M$ are related.

**Idea 2**: My guess is that $$\left(\sigma_{S^2}\circ\pi\right)(B)=\int_B\sigma_M({\rm d}y)\frac{\left|\langle\nu_M(y),\pi(y)\rangle\right|}{|y|^2}\tag4$$ (this would include to show that $\pi(B)\in\mathcal B(S^2)$). Since $\mathcal B(M)$ is generated by the open balls with center in $M$, it should be sufficient to prove $(4)$ for $B=B_\varepsilon(x)$ for some fixed $x\in M$ and $\varepsilon>0$. Now, $B$ is compact and has a $C^1$-boundary. Moreover, $$\nu_{\partial B}(y)=\frac{y-x}\varepsilon\;\;\;\text{for all }y\in\partial B\tag5$$ and $$\int\sigma_{\partial B}({\rm d}y)\frac{\left|\langle\nu_{\partial B}(y),\pi(y-x)\rangle\right|}{|y-x|^2}=\frac1{\varepsilon^2}\sigma_{\partial B}(\partial B)=\sigma_{S^2}(S^2)=\left(\sigma_{S^2}\circ\pi\right)(\partial B).\tag6$$ However, I don't know if $(6)$ is helpful for showing $(4)$. (But intuitively, the solid angle subtended by $\partial B$ should be the same as the solid angle subtended by $B$.)

EDIT: Isn't there a "visibility" function $v:M\times M\to\{0,1\}$ missing on the right-hand side of $(2)$ indicating whether $y$ is "visible" as seen from $x$ ($v(x,y)=1$) or is occluded by another point $z\in M$ with $\omega_{x\to y}=\omega_{x\to z}$ and $|x-z|<|x-y|$?

$^1$ i.e. $U\subseteq\mathbb R^2$ is open and $\psi:U\to M$ is an immersion and a topological embedding of $U$ onto $M$.

$^2$ Let $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$, $J_\phi$ denote the Jacobian of $\phi$ and $g_\phi:=\det J_\phi^TJ_\phi$. Then $$\sigma_M=\sqrt{g_\phi}\left.\lambda^{\otimes 2}\right|_U\circ\phi^{-1}.$$

$^3$ for example, [1, p. 14], [2, p. 53 (PDF numbering)] or [3, p. 6].

$^4$ i.e. for all $x\in\partial K$ there is an open neighborhood $V$ of $x$ and a continuously differentiable $\psi:V\to\mathbb R$ with $K\cap V=\{\psi\le0\}$ and $\nabla\psi\ne0$.

$^5$ i.e. if $\psi$ is as in footnote 4, then $$\nu_{\partial K}(x):=\frac{\nabla\psi(x)}{\left|\nabla\psi(x)\right\|}.$$