Let $U\subseteq\mathbb R^2$ be open, $\phi:U\to\mathbb R^3$ be an immersion and a topological embedding of $U$ onto $M:=\phi(U)$$^1$, $\nu_M(x)$ denote the unit normal vector of $M$ with $$\det\left({\rm D}\phi(u),\nu_M(x)\right)>0\tag1$$ for all $x\in M$ and $u=\phi^{-1}(x)$$^2$, $\sigma_M$ denote the surface measure on $M$$^3$ and $$\pi:\mathbb R^3\setminus\{0\}\to S^2\;,\;\;\;x\mapsto\frac x{|x|}$$ denote the projection of $\mathbb R^3\setminus\{0\}$ onto the unit 2-sphere $S^2$.
Assume $\left.\pi\right|_M$ is injective. How can we show that $$\sigma_{S^2}(B)=\int_{\pi^{-1}(B)}\sigma_M({\rm d}y)\frac{\left|\langle\nu_M(y),\pi(y)\rangle\right|}{|y|^2}\tag5$$ for all $B\in\mathcal B(S^2)$ with $B\subseteq\pi(M)$?
$^1$ which is to say that $M$ is a 2-dimensional embedded submanifold of $\mathbb R^3$ with global chart $\phi$.
$^2$ i.e. $$\nu_M\circ\phi=\frac{\partial_1\phi\times\partial_2\phi}{\left|\partial_1\phi\times\partial_2\phi\right|}\tag2.$$
$^3$ i.e. $$\sigma_M=\sqrt{g_\phi}\left.\lambda^2\right|_U\circ\phi^{-1}\tag3,$$ where $J_\phi$ is the Jacobian of $\phi$ and $$g_\phi:=\det\left(J_\phi^TJ_\phi\right)=\left|\partial_1\phi\times\partial_2\phi\right|^2\tag4.$$
One possible approach will be to compute over tiny patches in $M$ where the problem, up to arbitrary precision, reduces to assuming that $M$ is a plane with (constant) normal vector $\nu$. If a piece of a plane is projected onto $\mathbb{S}^2$ how much is the area? Once you express this as an integral over the (relevant part of the) plane, you simple sum over all pieces.
More on the reduction: Given $\epsilon >0$, locally, $M$ and all its geometric properties are order $\epsilon$-close to the corresponding properties of the tangent plane. Because the projection from $M$ to the tangent is almost an isometry, i.e. the Riemannian metric on $M$ and $T_pM$ are $\epsilon$-close to one another. The map is in particular almost $1$-Lipschitz.