No, itNow the statement is not trueindeed correct. What follows below is and answer to the previous version of the question which I'll keep for the moment.
Note that the condition for $f(C)$ to have non-zero winding number around $0$ is more-less empty. Indeed, to fix the winding number of $f(C)$ around $0$ one needs to choose infinity. And for more-less any closed curve $\eta$ in $S^2$ that doesn't pass through $0$ one can choose $\infty \in S^2$ so that the winding number of $\eta$ in $S^2\setminus \infty$ around $0$ is non-zero.
Now, for a concrete set of counter-examples suppose that $f: S^3\to S^2$ is ANY map with non-zero differential at a point $x\in S^3$ such that $f(x)\ne 0$. Then take a small ball $U$ containing $x$ such that $0\notin f(U)$ and a curve $\gamma\subset U$ that projects to a small circle in $S^2$. Choose $\infty\in S^2$ such that $f(\gamma)$ separates $0$ from $\infty$. Then clearly the winding number of $f(\gamma)$ around $0$ is $\pm 1$ but $f^{-1}(0)$ doesn't contain a component linked to $c$.
I can see only one way to fix this. Ask $f$ not to be null-homotopic and ask $C$ to be a full premiage of a point $x\in S^2$ different from $0$ ...