How to derive that after stereo-graphical projection, $\Delta u$ in $\mathbb{R}^n$ is transformed to $$ \Delta_{\mathbb{S}^n}u - \frac{n(n-2)}{4}u\ \text{in}\ \mathbb{S}^n. $$ To be more precise, in this paper, it is asserted that the equation $$-\Delta_{\mathbb{S}^n}\,u+\frac{n(n-2)}{2}u-\tilde{K}u^{\frac{n+2}{n-2}}=0\quad\text{on}\ \mathbb{S}^n$$ can be reduced to $$-\Delta u = K(y)u^{\frac{n+2}{n-2}}\quad\text{in}\ \mathbb{R}^n.$$
I need some tips or relevant references.
the relation between the two Laplacians is a bit more complicated. I follow the survey paper of Lee and Parker on the Yamabe problem.
The conformal Laplacian of a Riemannian metric $g$ is defined as $$ L_g = 4 \frac{n-1}{n-2} \Delta_g + S_g $$ where $S_g$ is the scalar curvature of the metric and $\Delta_g = d_g^* d$ is the Hodge (positive) Laplacian. The main point is that if we do a conformal change of the metric $\tilde{g} = \varphi^2 g$ (where $\varphi>0$ is a smooth positive function) then \begin{equation}\label{eq:conflap} L_g (u) = \varphi^{\frac{n+2}{2}} L_{\tilde{g}} (\varphi^{\frac{2-n}{2}} u) . \end{equation} The metric of the unit sphere in stereographic projection coordinates is \begin{equation*} g_{S^n} = \varphi^2 g_{\mathbb{R}^n} \hspace{2mm} \text{ with } \varphi = \frac{2}{(1+|x|^2)}. \end{equation*} Using the equation for the conformal laplacian and that the scalar curvature of $S^n$ is $n(n-1)$ we obtain \begin{equation} \Delta_{\mathbb{R}^n} u = \varphi^{\frac{n+2}{2}} \left( \Delta_{S^n} + \frac{n(n-2)}{4} \right) (\varphi^{\frac{2-n}{2}} u) . \end{equation}
On the other hand, the conformal Laplacian relates to prescribing scalar curvature in conformal classes as follows. Let $K$ be a given function, then the metric $\tilde{g} = u^{4/(n-2)} g$ has scalar curvature equal to $K$ if and only if $u$ solves the PDE \begin{equation*} L_g (u) = u^{\frac{n+2}{n-2}} K . \end{equation*} In the case of the sphere, the metric $\tilde{g} = u^{4/(n-2)} g_{S^n}$ has scalar curvature equal to $K$ if and only if \begin{equation}\label{eq1} \Delta_{S^n} u + \frac{n(n-2)}{4} u = u^{\frac{n+2}{n-2}} \tilde{K} \end{equation} where $\tilde{K}= \frac{n-2}{4(n-1)} K$. At the same time, $\tilde{g}= \tilde{u}^{4/(n-2)} g_{\mathbb{R}^n}$ where $\tilde{u} = \varphi^{\frac{n-2}{2}} u$ with $\varphi = 2 (1+|x|^2)^{-1}$. We conclude that $u$ solves the above equation if and only if $\tilde{u}$ solves
\begin{equation} \Delta_{\mathbb{R}^n} \tilde{u} = \tilde{u}^{\frac{n+2}{n-2}} \tilde{K} . \end{equation}
