Let $z\in S^2$. Fix vector $v\in S^2$. Then we can write $z=\langle z,v\rangle v+ \sqrt{\left(1- \langle z,v\rangle^2\right ) } v^{ \perp} $ where $v^{\perp}\in S^2$ such that $v$ and $v^{\perp}$ are orthogonal. Let $a= \langle z,v\rangle $
We need to change variable from $z\to (a,v^{ \perp})$, then $dz= J(a,v^{\perp})dadv^{\perp}$. I am interested in finding the jacobian factor $J$. Is there any way to find that? I tried hard. But I could not get it.
Any help or hint will be highly appreciated.
Thanks
Let me work with Euclidean coordinates, $$ z=(x_1,x_2,x_3)\in S^2. $$
There certainly exists an orthogonal transformation $L:\mathbb R^3\to\mathbb R^3$ such that $Lv=(1,0,0)$. Fix such an $L$ and denote $Lz=(x_1',x_2',x_3')$.
The coordinate transformation is $z\mapsto (a,t)$ where $$ a=\langle z,v\rangle=\langle Lz,Lv\rangle = x'_1 $$ and $$ (\cos t,\sin t)=\frac 1{\sqrt{1-a^2}}(x_2',x_3')=v^\perp. $$ (Whenever $a\not=1$, i.e. for all $z\not=v$.)
Let us use also the spherical coordinates $$ x_1'=\cos\theta,\ \ x_2'=\sin\theta\cos\phi,\ \ x_3'=\sin\theta\sin\phi. $$ Hence, $a=\cos\theta$, $\sin\theta\mathrm d\theta=\mathrm da$, $t=\phi$. The area form $\mathrm dz$ is $$ \mathrm dz = \sin\theta \mathrm d\phi\,\mathrm d\theta = \mathrm da\, \mathrm dt. $$ (Here we use that $L$ is an isometry of $\mathbb R^3$ hence it preserves the area form on the sphere $S^2$.)
The coordinates you are describing are just rotated spherical coordinates. I am using the coordinate $t$ to describe the vector $v^\perp$.