Prove this vector field on $S^2$ is smooth I am having solving the following exercise:

(iii) Fix the coordinate chart $(U_1, f_1)$ on $S^2 = \{(x,y,z) \in
 \mathbb{R}^3 : x^2 + y^2 + z^2 = 1 \} \subset \mathbb{R}^3$ where $U_1
 = S^2 \cap \{z>0\}$ and $f_1 : U_1 \to \mathbb{B}^2 = \{(s,t) : s^2 + t^2 = 1\} \subset \mathbb{R}^2$ is defined as $f_1(x,y,z)=(x,y)$.
Consider $T_{p} S^2$ as a subspace of $\mathbb{R}^3$ for every point
$p$ of $S^2$. Prove that the following vector field on $S^2$: $$ \zeta
 : S^2 \to TS^2$$ $$ (x,y,z) \mapsto ((x,y,z),(xz,yz,z^2-1))$$ is
smooth on the open subset $U_1 \subset S^2$.

Part (i) of the question (which I believe is relevant) asks us to prove the following:

Fix a point $p = (x,y,z) \in U_1$. Find two smooth curves $\sigma_1,
 \sigma_2$ in $S^2$ through the point p such that they are not tangent
at $p$. Consider the linear isomorphism $$ \Delta_{f_1} : T_{p}S^{2}
 \to \mathbb{R}^2$$ which sends the equivalence class of a curve
$\sigma$ through p to $D(f_1 \circ \sigma)|_0 $. Find
$\Delta_{f_1}([\sigma_1])$ and $\Delta_{f_1}([\sigma_2])$

My trouble is with the first exercise not the second.
My attempt: 
We essentially are trying to show a map between manifolds is smooth, so we fix a point $p \in U_1 \subset S^2$. Given the chart $(U_1, f_1)$ containing $p$ we need to show that $g \circ \zeta \circ f_{1}^{-1} : f_{1}(U_{1}) \to g(V)$ is smooth, where $(V, g)$ is a chart on $T_p S^2$ containing $\zeta(p)$. But here I get stuck because I don't know what to do with $g$; I can compute $\zeta \circ f_{1}^{-1}$ but how can I represent $g$?
My questions:

*

*Is my initial approach correct? If not, how would we solve the exercise?

*What does it really mean to "Consider $T_{p} S^2$ as a subspace of $\mathbb{R}^3$ for every point"?

Thank you very much.
 A: First, a note on the definition of $f$: it is the projection from the upper hemisphere to the disk, so $\mathbb{B}$ should be $\{(x,y)\mid x^2+y^2<1\}$, rather than $\{(x,y)\mid x^2+y^2=1\}$. Anyway, a vector field on $\mathbb{S}^2$ is smooth if and only if its coefficients in a system of coordinates are smooth. So we should write the vector field $\zeta$ in the coordinates $(x,y)$ for $\mathbb{S}^2$ given by $f_1$, and check that the coefficients are smooth. At least, this is how I would approach the exercise. Below, you can find my solution, but try it yourself first!
It might be confusing to use the same symbols both for the coordinates on $\mathbb{R}^3$ and $\mathbb{B}$, so let's use $(u,v)$ as coordinates $\mathbb{B}$. The inverse of $f_1$ is given by $$(u,v)\mapsto\left(u,v,\sqrt{1-(u^2+v^2)}\right).$$
We are considering the vector field $\zeta$ as a map $\mathbb{S}^2\to\mathbb{R}^3$, this is what the exercise means by saying that we are considering $T_p\mathbb{S}^2$ as a subset of $T_p\mathbb{R}^3=\mathbb{R}^3$. With this in mind, we have $$\zeta(x,y,z)=xz\,\partial_x+yz\,\partial_y+(z^2-1)\partial_z.$$
It is easy to check that $\zeta\in T_{(x,y,z)}\mathbb{S}^2$, that is $$(x,y,z)\cdot(xz,yz,z^2-1)=0$$
for every $(x,y,z)\in\mathbb{S}^2$.
Now we should express the vector field $\zeta$ in the coordinate system $(u,v)$. Computing the differential of the map $f_1^{-1}:\mathbb{B}\to\mathbb{R}^3$ we get
\begin{equation}
\begin{split}
\partial_u=&\partial_ux\partial_x+\partial_uy\partial_y+\partial_uz\partial_z=\\
=&\partial_x-\frac{u}{\sqrt{1-(u^2+v^2)}}\partial_z\\
\partial_v=&\partial_vx\partial_x+\partial_vy\partial_y+\partial_vz\partial_z=\\
=&\partial_y-\frac{v}{\sqrt{1-(u^2+v^2)}}\partial_z.
\end{split}
\end{equation}
At this point a quick computation gives (setting $r^2=u^2+v^2$)
\begin{equation}
\begin{split}
\zeta(u,v)=&\sqrt{1-r^2}\left(u\partial_x+v\partial_y\right)-r^2\partial_z=\\
=&\sqrt{1-r^2}\left(u\partial_u+\frac{u^2}{\sqrt{1-r^2}}\partial_z+v\partial_v+\frac{v^2}{\sqrt{1-r^2}}\partial_z\right)-r^2\partial_z=\\
=&\sqrt{1-r^2}\left(u\partial_u+v\partial_v\right).
\end{split}
\end{equation}
As $r^2<1$ for $(u,v)\in\mathbb{B}$, the coefficients of $\zeta$ in this coordinate system are smooth.
