Let
$\overline{S}^n_K$ be the stereographic projection of the sphere, and
$S^n_K$ be the sphere with sectional curvature $K$.
Let $\Pi$ be the stereographic projection $$ \Pi\colon S^n_K\setminus\{\text{north pole}\}\to\overline{S}_K^n. $$
Then, for example, the distance function on $\overline{S}^n_K$ can be obtained by back-projecting the points to $S^n_K$ and inserting them into the distance formula on $S^n_K$.
$$ d_{\overline{S}^n_K}(x,y)=d_{S^n_K}(\Pi^{-1}(x), \Pi^{-1}(y)). $$
This formula can then be simplified and gives a distance function on $\overline{S}^n_K$.
I know that the exponential map on $S_K^n$ is
$$ \begin{align*} \exp_{x}\colon T_{x}S^n_K&\to S^n_K\\ u&\mapsto \cos\left( \sqrt{K}||u||_{x} \right)x + \frac{ \sin\left( \sqrt{K}||u||_{x} \right) }{ \sqrt{K}||u||_{x} } u. \end{align*} $$
Now what I'd like to do is to derive is the exponential map on $\overline{S}^n_K$ in a similar fashion.
That means I'd have to back-project $x$ with $\Pi^{-1}$.
Now, how do I transform the tangent vector $u'\in T_x \overline{S}^n_K$ to get the corresponding tangent vector in $u\in T_{\Pi^{-1}(x)}S_K^n$?
$$ u'\in T_x \overline{S}^n_K \quad \to \quad u\in T_{\Pi^{-1}(x)} S^n_K $$