Vector fields on $\Bbb S^2$ with zero construct a vector field on ${S} ^ 2$ with only one zero?
 $$\mathbb{S}^2= \left\{(x_1,x_2,x_3) \in R^3: x_1^2+x_2^2+x_3^2=1\right\}$$
 A: Write down the stereographic projection from the plane $R^2$ to the sphere, with projection from the north pole. Call that F.
For each $(x, y)$ in $R^2$, take 
$$
DF(x, y) \begin{bmatrix} 1/(1 + x^2 + y^2) \\ 0 \end{bmatrix}
$$
That will be a vector at $F(x, y) \in S^2$. It'll be nonzero everywhere except the north pole. Fill in with the zero vector at the north pole. 
Addition in response to comment: stereographic projection takes a point $P$ of the plane and joins it by a line to the north pole, $N = (0,0,1)$ of the sphere. That line meets the sphere at another point $F(P)$, called "the stereographic projection of $P$". In coordinates, it's the map
$$
(x, y) \mapsto \frac{1}{x^2 + y^2 + 1} \left( 2x, 2y, x^2 + y^2 - 1 \right).
$$
Added after further work:
Let's use $uv$ coordinates on the plane, $xyz$-coordinates on the sphere. Then projection from the plane to the sphere is 
$$
P : R^2 \to S^2 : (u,v) \mapsto \frac{1}{u^2 + v^2 + 1} (2u, 2v, u^2 + v^2 -1)
$$
while projection from the sphere to the plane is 
$$
H : S^2 \to R^2 : (x, y, z) \mapsto (\frac{x}{1-z}, \frac{y}{1-z}). 
$$
The functions $H$ and $S$ are inverses of one another. 
Letting $r$ denote $u^2 + v^2 + 1$ to keep things a little shorter, the derivative of $P$ at $(u, v)$ is 
\begin{align}
DP(u, v) &= \begin{bmatrix} 
\frac{2r - 4u^2}{r^2} & \frac{-4uv}{r^2} \\
\frac{-4uv}{r^2} & \frac{2r - 4v^2}{r^2}\\
\frac{2ur - 2u(u^2 + v^2 - 1)}{r^2} & \frac{2vr - 2v(u^2 + v^2 - 1)}{r^2} 
\end{bmatrix}\\
 &= \begin{bmatrix} 
\frac{2r - 4u^2}{r^2} & \frac{-4uv}{r^2} \\
\frac{-4uv}{r^2} & \frac{2r - 4v^2}{r^2}\\
\frac{4u}{r^2} & \frac{4v}{r^2} 
\end{bmatrix}
\end{align}
If we multiply this by the vector $\begin{bmatrix} 1\\0 \end{bmatrix}$, we'll get a tangent vector to $S^2$ at the point $P(u, v)$, namely
\begin{align}
V_1 &= \begin{bmatrix} 
\frac{2r - 4u^2}{r^2}  \\
\frac{-4uv}{r^2} \\
\frac{4u}{r^2} 
\end{bmatrix}
\end{align}
which is proportional to 
\begin{align}
V_2 &= \begin{bmatrix} 
{2r - 4u^2}  \\
{-4uv} \\
{4u} 
\end{bmatrix}
\end{align}
It'd be nice, though, to have that in terms of the sphere point $(x, y, z)$ rather than its $uv$ parameters. Using the substitution provided by $H$, we can replace $u$ and $v$ by $x/(1-z)$ and $y/(1-z)$ and simplify (using $x^2 + y^2 + z^2 = 1$), and scaling up or down as needed to simplify the algebra. We eventually get a vector field $V_3(x, y, z)$ which is proportional to $V_2$, and nonzero everywhere except the north pole, namely
$$
V_3(x, y, z) = \begin{bmatrix} 
x^2 + z - 1 \\
xy \\
x(z-1)
\end{bmatrix}.
$$
How do we know that $V_3$ is nonzero except at the poles? Well, the middle entry is nonzero only if $x = 0$ or $y = 0$, i.e., the union of two great circles, each of which passes through the north pole. On the circle $x = 0$, the first term is $-1 + z$, which is zero only when $z = 1$, i.e., at the north pole. On the circle $y = 0$, we have $x \ne 0$ (except at the poles), and $z \ne 1$ (except at the poles), so the third term is nonzero everywhere except the poles. 
To summarize: on the union of the circles $x= 0$ and $y = 0$, ignoring the north pole, the first entry of the field is nonzero on $x = 0$, while the third is nonzero on $y = 0$, except at the south pole (at which the first entry is nonzero). In other words, at least one entry of the field is nonzero everywhere except the north pole. At the north pole, all three entries are zero, and the morse index theorem tells us the index must be -2. 
Alternatively, you can compute the squared length of the vector $V_2(x, y, z)$ and find that it's $(z-1)^2$, which is evidently nonzero everywhere on the sphere except the north pole. 
