I will answer your question for $S^{2n+1}$, since there is no difference between the case $n=1$ and the case of general $n$.

Let $(M^{2n+1},g,\theta)$ be a Sasakian manifold.  One definition of a Sasakian manifold is that its metric cone is Kähler; this is the "one above".  Here the *metric cone* is the manifold $M\times(0,\infty)$ with metric $dt^2+t^2g$.  Thus in the case of $S^{2n+1}$, the metric cone is $\mathbb{C}^{n+1}$ with the flat metric (written in spherical coordinates).

Taking the quotient by the $S^1$-action on $(M^{2n+1},\theta)$ determined by the Reeb vector field gives the "Kähler manifold below".  For $S^{2n+1}$, the $S^1$-action is scalar multiplication by $e^{i\phi}$ (regarding $S^{2n+1}$ as the unit sphere in $\mathbb{C}^{n+1}$), so the quotient is $\mathbb{C}P^n$.