Let $K$ be a compact subset $\mathbb R^3$ with a $C^1$-boundary, i.e. for all $x\in\partial K$ there is an open neighborhood $U$ of $x$ and a $\psi\in C^1(U)$ with $$K\cap U=\left\{u\in U:\psi(u)\le0\right\}\tag1$$ and $$\psi'(u)\ne0\;\;\;\text{for all }u\in U.$$ Let $\sigma_{\partial K}$ and $\nu_{\partial K}$ denote the surface measure and outer normal field on $\partial K$, respectively. Moreover, let $$\pi:\mathbb R^3\setminus\{0\}\to S^2\;,\;\;\;x\mapsto\frac x{|x|}$$ denote the projection onto the unit 2-sphere $S^2$ and $\sigma_{S^2}$ denote the surface measure on $S^2$.

From geometric inspection (see, for example, page 15 of https://backend.orbit.dtu.dk/ws/portalfiles/portal/51112612/phd240_awk_vers2.pdf and the picture below), we should have $$\sigma_{S^2}({\rm d}\omega_{x\to y})=\sigma_{\partial K}({\rm d}x)\frac{\left|\langle\nu_{\partial K}(x),\omega_{x\to y})\rangle\right|}{|x-y|^2}\tag2,$$ where $$\omega_{x\to y}:=\pi(y-x)\;\;\;\text{for }x,y\in\mathbb R^3\text{ with }x\ne y,$$ but I struggle to state and prove this in a measure-theoretic rigorous way.

If $f(x):=|x|$ for $x\in\mathbb R^3$, we may note that $$\langle\nu_{\partial B},\pi\rangle=\langle\nu_{\partial B},\nabla f\rangle=:\frac{\partial f}{\partial\nu_{\partial K}}\tag3$$ and hence this result may be related to the divergence theorem.

So, how do we need to understand $(2)$ in the language of measure theory and how can we prove it?

**EDIT**: Another approach could be to consider the open balls in $K$ since they generate $\mathcal B(K)$. The left-hand side is then easy to compute: If $x\in\mathbb R^3$ and $\varepsilon>0$, $$\sigma_{S^2}\left(\pi(B_\varepsilon(x))\right)=\sigma_{S^2}(S^2)=4\pi.\tag4$$ Now we may consider the right-hand side over $B_\varepsilon(x)$ (assuming this ball is contained in $K$).

(In the picture, $A=\sigma_{\partial K}$, ${\rm n}=\nu_{\partial K}$ and $\sigma=\sigma_{S^2}$.)