The change of coordinates from north pole coordinates to south pole oordinates is $\phi:\mathbb R^2\setminus \lbrace (0,0)\rbrace \to \mathbb R^2\setminus \lbrace (0,0)\rbrace:\begin{pmatrix} x \\\\ y\end{pmatrix}\mapsto \begin{pmatrix} u \\\\ v\end{pmatrix}=\frac {1}{(x^2+y^2)}\begin{pmatrix} x \\\\ y\end{pmatrix}$ and the jacobian of that map is given by the matrix
$Jac(\phi)(x,y)=\frac {1}{(x^2+y^2)^2}\begin{pmatrix} -x^2+y^2&-2xy \\\\ -2xy&x^2-y^2 \end{pmatrix}$
This implies that your vector field has as expression in the south pole chart(beware the minus sign!) $Y(u,v)=-(u\partial_u+v\partial_v)\quad \text{for} \begin{pmatrix} u \\\\ v\end{pmatrix}\in \mathbb R^2\setminus \lbrace (0,0)\rbrace $ which can be smoothly extended to the origin by $Y(0,0)=0\partial_u+0\partial_v$.
Conclusion
The initial vector field $X$ on the sphere minus the north pole smoothly extends to a global vector field on the whole sphere by defining it as zero at the north pole.
Edit
Slightly more generally we can define a global vector field (vanishing at the two poles and only there if the real numbers $a,b$ are not both zeroon the sphere by the following expressions in the two charts given by the two stereographic projections: $(ax+by)\partial_x+(-bx+ay) \partial_y=(-au+bv)\partial_u+(-bu-av)\partial_v$ If the real numbers $a,b $ are not both zero, that vector field vanishes at the two poles and only there.