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, consisting of $N$ continuous segments $z_1(t),z_2(t),\ldots z_N(t)$, connected by $\pm 2\pi$ jumps at each $t=t_1,t_2,\ldots t_{N-1}$ where the branch cut of the logarithm is crossed:
$$z_p(t_{p})-z_{p+1}(t_p)=\sigma_p 2\pi,\;\;p=1,\ldots N-1,\;\;\sigma_p\in\{+1,-1\}.$$ 
We define $t_0=0$, $t_N=1$. The integral then evaluates to
$$\int_0^1 \frac{dz(t)}{dt}\,dt=\sum_{p=1}^N \int_{t_{p-1}}^{t_p}\frac{dz_p(t)}{dt}\,dt=\sum_{p=1}^N \bigl(z_p(t_p)-z_p(t_{p-1})\bigr)$$
$$\qquad=z_N(t_N)-z_1(t_0)+\sum_{p=1}^{N-1}\sigma_p 2\pi=\sum_{p=1}^{N-1}\sigma_p 2\pi,$$
because $z_N(t_N)-z_1(t_0)=z(1)-z(0)=0$.
The net sum of the $\pm 2\pi$ jumps is the winding number $W=\sum_{p=1}^{N-1}\sigma_p 2\pi$.

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.

<IMG SRC="https://i.sstatic.net/cKxgU.png" WIDTH="400"/>