Does there exist a smooth vector field from $S^3$ to its tangent bundle with exactly $3$ zeros?
The only vector field I can think was $v(x_1,x_2,x_3,x_4)=(-x_2,x_1,-x_4,x_3)$ which has no zeros. How could I think about a one that has zeros? Any help is appreciated.
For any vector field $v$ on any $n$-manifold $M$, for any point $p \in M$, for any neighborhood $U$ of $p$, and for any homeomorphism $f : B^n \to U$ where $B^n$ is the unit open ball in $\mathbb{R}^n$ and $f(0)=p$, you can alter the vector field $v$, leaving it unchanged on $M-U$, and creating a zero at $p$ with no other zeroes in $U$. Simply choose a non-negative smooth function $g : B^n \to \mathbb{R}$ such that $g(0)=0$, $g$ has no other zeroes in $B^n$, and $g$ is equal to $1$ for all $x \in B^n$ such that $|x| \ge 1/2$. Now scalar multiply the vector field by the smooth function which equals $g \circ f^{-1}$ on $U$ and equals $1$ on $M-U$.
To get your desired example from the vector field $v$ described in your question, simply choose three points, surround them by three pairwise disjoint open balls, and alter the vector field on those balls just as described above.