Tangent vector field on $S^2$ seen as a linear derivation on $C^\infty$ functions.

Question

I have to show that the map $Y(x,y,z)=(xz,yz,z^2-1)$ is a field vector on $S^2$. Intuitively it makes sense :
for each $(x,y,z) \in S^2$, $$\langle(x,y,z),Y(x,y,z)\rangle = x^2 z + y^2 z + z(z^2 -1) = z(x^2 + y^2 + z^2 -1) = 0$$ so $Y$ sends each point on $S^2$ to a normal vector to that point, i.e. a tanget vector on the sphere in that point. But formally we define a tangent vector as a derivation on the germs of $C^{\infty}$ functions in a neighborhood of the point. Fixed a point $p=(x,y,z) \in S^2$ how can I see $Y(x,y,z)$ as a derivation?
Fixed a germ of $f \in C^\infty(U)$ where $p \in U$, how can i define $Y(x,y,z)(f)$ ?
EDIT : Is it correct to define, for each local chart $(x_1,x_2,x_3)$, $$ Y(p)(f)= xz(\frac\partial{\partial x_1}f)_p + yz(\frac\partial{\partial x_2}f)_p + (z^2 -1)(\frac\partial{\partial x_3}f)_p $$
and see $Y$ as the field vector $xz\frac\partial{\partial x_1} + yz\frac\partial{\partial x_2} + (z^2 -1)\frac\partial{\partial x_3} $ ?