Let $u$ be a function defined on the unit sphere $S$ centered at $(0,0,-z_0)$, and $\Pi$ be the stereographic projection of $S$ onto $\mathbb{R}^2$. Indeed the part of the sphere that lies below $xy$-plane is mapped to a disc or radius $R=1-|z_0|^2$, and the upper half of the sphere is mapped to $\mathbb{R} \backslash B(0,R)$. Define $\bar{u}:\mathbb{R}^2 \rightarrow R$ by $\bar{u}(x)=u(\Pi^{-1}(x))$.
I wonder how the Laplacian of $u$ and $\bar{u}$ are related. In general one expects that $$\Delta\bar{u}(x)=H(x)\Delta u(\Pi^{-1}(x)),$$ for some function $H$.
I am having difficulties computing $H$. In the case of the standard stereographic projection ($R=1$) we have $$H=(\frac{2}{1+|x|^2})^2.$$
How does this generalize to the more general stereographic projection above?
Notation. The letters $S$ and $P$ are used as subscripts to label quantities that live on the sphere and plane respectively.
2D case with projection plane tangent to unit sphere. As a warm-up to case you posed, consider first the familiar case where the projection plane $P$ is tangent to the unit sphere $S$, and the point of tangency is the South Pole located at height $z=-1$. The projection map is a conformal map, and small distances on the respective surfaces are related by an isotropic dilation factor:
(1) $ds_S= \lambda ds_P$
where $ \lambda= \frac{2}{1+ |x|^2} \leq 1$. (This factor shrinks Euclidean planar distance so it correctly accounts for the smaller distances one sees on a sphere.)
If the sphere and plane both have dimension $n=2$, it turns out that $\triangle _S = \lambda^{-2} \triangle_P$ (See appendix).
2D case with arbitrary projection plane. If a new projection plane $P'$ is located at a different height $z=-H$, then each ray from the North Pole pierces the new plane at a point $x'$ and pierces the old tangent plane at a point $x$, and these two planar points satisfy the proportionality relation $ x'= k x$ where $k=\frac{1+H}{2}$ is a constant factor. It follows that likewise distances are related by $ds_{P'} = k ds _P$. Thus
(2) $ds_S= \lambda ds_P= \frac{\lambda}{k} ds_{P'} = \lambda' ds_{P'}$
where $\lambda'= \frac {\lambda}{k} = \frac{2}{ k(1+ (\frac{ |x'|}{k})^2} = \frac{ 2k}{ k^2+ |x'|^2}$. .
Finally, $\triangle_S = \lambda'^{-2} \triangle_{P'} $
Appendix. Derivation of conversion formula for conformally related Laplacians in arbitrary dimensions
Because all distances are locally proportional it is easy to derive the conversion formula that relates their respective Laplacians.Your query is in regard to the case $n=2$ but the same ideas apply in arbitrary dimensions.)
Note $ grad_S f$, which is a rate of change of $f$ per unit length, is $ \frac{d f}{ds_S} = \lambda^{-1} grad_Pf$.
Now, $ \triangle _S f = div_S grad_S f$ satisfies the divergence theorem relation $ \triangle _S f dV_S $= (flux of $ grad_S f$ across a small cell on $S$)/ (volume of that small cell on S)
The boundary of each $(n-1)-$ dimensional cell on the sphere has "surface area" $ dA_S= \lambda^{n-1} dA_P$ and the volume of that spherical cell is $dV_S= \lambda^n dV_P$
Thus $\triangle _S f= \lambda^{-n} div_P[ \lambda^{n-1} (\lambda^{-1} grad_Pf)] = \lambda^{-n} div_P ( \lambda^{n-2} grad_P f)$.
This expands out to give $\triangle _S f= \lambda^{-2} [\triangle_P f + (n-2) \frac{ grad_P \lambda}{\lambda} \cdot grad f]$.
On a sphere $S'$ of radius $R>1$ the same method works. You can reduce to the case treated above. You just need to scale up the unit sphere and and its tangent plane by the factor $R$ to get the bigger sphere $S'$. After some calculation you will get $ \lambda' = \frac{ 2 }{1+ |x|^2/R^2}$.
This follows from the conversion rules $ds_{S'} = R ds_S $ and $ ds_P' = R ds_P$ and $x'/R =x$