Can this statement about winding number generalized? 
Definition
Let $\alpha$ be a path in $\mathbb{C}\setminus\{z_0\}$.
Since $\mathbb{C}\rightarrow \mathbb{C}\setminus \{z_0\}:z\mapsto e^z$ is a covering map, $\alpha$ can be decomposed as $\alpha=z_0+ re^{i\theta}$ where $r>0$ and $\theta$ is real.
Define $Wnd(\alpha,z_0)=\frac{\theta(1)-\theta(0)}{2\pi}$.
Note that this definition does not depend on a choice of $\theta$.

Under this definition, I have proven the following:

Let $f,g:S^1\rightarrow \mathbb{C}\setminus\{0\}$ be continuous functions and $x_0\in S^1$.
Let $\alpha$ be a loop at $x_0$ in $S^1$ such that $[\alpha]$ is not the identity in $\pi_1(S^1,x_0)$.
Then, $f,g$ are homotopic iff $Wnd(f\circ \alpha,0)=Wnd(g\circ \alpha,0)$.

I proved it very indirectly, so I'm not sure whether this can be generalized further. That is, even if $\alpha$ is not a loop ($\alpha(0)\neq \alpha(1)$), does the statement still hold?
If not, what is a counterexample?
I think $\Rightarrow$ direction is likely to hold, but $\Leftarrow$ direction might be false.

For non-loops, neither implication holds. If we take the half-circle $\alpha \colon [0,1] \to S^1,\: \alpha(t) = e^{\pi i t}$, and look at $f(z) = z$,
$$g = f\ast c_1 \colon z \mapsto \begin{cases} z^2 &, \operatorname{Im} z \geqslant 0 \\ 1 &, \operatorname{Im} z < 0,\end{cases}$$
and $h(z) = z^2$, then we have $f \simeq g \not\simeq h$, but $g\circ\alpha = h\circ\alpha$. From that we obtain
$$\operatorname{Wnd}(g\circ\alpha,0) = \operatorname{Wnd}(h\circ\alpha,0) = 1,$$
whereas $\operatorname{Wnd}(f\circ\alpha,0) = \operatorname{Wnd}(\alpha,0) = \frac{1}{2}$.
It doesn't help if we require that $\lvert\operatorname{Wnd}(\alpha,0)\rvert \geqslant 1$: for $\alpha(t) = e^{3\pi i t}$, we have with the functions $f,g \colon S^1\to S^1$ from above $\operatorname{Wnd}(f\circ\alpha,0) = \frac{3}{2} \neq 2 = \operatorname{Wnd}(g\circ\alpha,0)$. And for
$$h(z) = \begin{cases} z^6 &, \operatorname{Im} z \geqslant 0 \\ 1 &, \operatorname{Im} z < 0\end{cases}$$
and $k(z) = z^4$ we have
$$\operatorname{Wnd}(h\circ\alpha,0) = 6 = \operatorname{Wnd}(k\circ\alpha,0),$$
while $h \simeq (z\mapsto z^3) \not\simeq k$.
