Relate the solid angle and surface measure of a surface 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 $$\nu_{\partial K}(x):=\frac{\nabla\psi(x)}{\left|\nabla\psi(x)\right\|}.$$
 A: The density you are asking about is precisely the Jacobian of the central projection of $M$ to the unit sphere centered at the origin (as was already mentioned by Willie Wong). For an explicit expression (for instance, in the form given on p.14 in the reference you quote) it is enough to look at the picture on p.15 (which is apparently what you mean by "geometric inspection"). If you wish, one can do it in two steps by first replacing $M$ with the corresponding tangent plane (which does not change the Jacobian). I really fail to see what is the "measure-theoretic rigor" one needs in addition to this observation (which does not really seem to be at the research level). However, if you want a mathematical reference, this formula is, for instance, derived in great detail (in a somewhat different notation though) in Palamodov's book Reconstructive integral geometry, (Section 3.1, Example 5).
Talking about notation, the way you reproduce the (already somewhat confused) formula from p.14 (not p.15 as you write) of your reference is completely erroneous (which was also mentioned by Willie Wong). In that reference a point $y\in M$ is projected to the unite sphere centered at $x$, whereas in your notation it is $x\in M$ that is projected to the unite sphere centered at 0 (and the role of $y$ is completely unclear).  
The right expression for the density is $|\langle x, \nu_M(x) \rangle|/|x|^3$. Geometrically, the numerator is the distance from the origin to the tangent plane to $M$ at the point $x$.
