Let me work this out, since the comments do not seem to have resolved the issue.

The function $f$ has a root at $x=0=y$ so I integrate along a circle centered at the origin, $x+iy=\varepsilon\cos 2\pi t+i\varepsilon\sin 2\pi t$. I decompose into modulus and phase,
$$f(x,y)=|f(t)|e^{i\phi(t)},$$
with $|f(0)|=|f(1)|$ and $\phi(0)=\phi(1)$. The logarithm has a branch cut along the negative real axis, which is not crossed by $|f(t)$, hence
$$\int_0^1 \frac{d\ln |f(t)|}{dt}\,dt=\ln|f(1)|-\ln|f(0)|=0.$$
The function $z(t)=-i\ln e^{i\phi(t)}\in(-\pi,\pi]$ is a piecewise continuous function, with continuous segments connected by $\pm 2\pi$ jumps. The contribution to the integral from the continuous segments vanishes, while the discontinuities contribute $\pm 2\pi$. The net sum of the $\pm 2\pi$ jumps is the winding number.

In these plots I compare $z(t)$ for the two functions mentioned by the OP, $f_1=x+iy$ (left plot) and $f_2=x+iy^2$ (right plot). So you see that the first function has one $2\pi$ jump, winding number 1, while the second function has no jump at all, winding number 0.

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