This can be done in four steps:
1) Compute linear transformation (T) that maps triangle A to its counterpart B on the sphere.
2) Use T to map point of interest Pa in A to B to get Pb=T(Pa)
3) Project Pb onto the sphere to get pb=Pb/norm(Pb)
4) Solve for spherical barycentric coordinates (u,v,w) of pb relative to B. Do to this suppose Q=[q1 q2 q3] is a 3-by-3 matrix containing coordinates of the vertices of B along columns, then p2(:)=Q*[u;v;w]. Note that unlike planar barycentric coordinates which always sum to unity, sum of the spherical barycentric coordinates can exceed one.