Why does the winding number become zero when the region is unbounded

As you formulate it, your question "Why does the winding number become zero when the region is unbounded" does not make much sense. The path integral $$\frac1{2\pi i}\int_\gamma\frac{\mathrm dz}{z-a}$$ is defined for any closed curve $\gamma : [r,s] \to \mathbb C$ such that $a$ is not contained in the image of $\gamma$. You do not need any region. And in fact the value of this integral is a number in $\mathbb Z$ (which definitely depends on the point $a$). Thus the statement $$\lim_{a\to\infty}\left|\frac1{2\pi i}\int_\gamma\frac{\mathrm dz}{z-a}\right| = 0$$ makes sense for any closed curve. For the limit the absolute value is irrelevant, it has the same meaning as $$\lim_{a\to\infty}\frac1{2\pi i}\int_\gamma\frac{\mathrm dz}{z-a}\ = 0 .$$ In other words, we have to show $$I(a) = \int_\gamma\frac{\mathrm dz}{z-a} = \int_r^s\frac{\gamma'(t)}{\gamma(t)-a}dt = 0$$ for $\lvert a \rvert > R$.

The image of $\gamma$ is contained in an open disk with center $0$ and sufficiently large radius $R$. Let $\bar \gamma(t) = \gamma(t) - a$. Then $$I(a) = \int_r^s \frac{\bar \gamma'(t)}{\bar\gamma(t)}dt.$$ If $\lvert a \rvert > R$, then the image of $\bar \gamma$ is contained in an open disk $D$ with center $a$ and radius $R$. This disk does not contain $0$, thus there exists a branch $\ln$ of the complex logarithm on $D$. We get $$I(a) = \int_r^s \ln'(\bar\gamma(t))dt = \ln(\bar\gamma(s)) - \ln(\bar\gamma(r)) = 0 . $$

Remark:

The above result implies that the winding number of $\gamma$ is $0$ on the unbounded component of $\mathbb C \setminus \gamma([r,s])$. Note that there is exactly one unbounded component; it contains all points with absolute value $> R$.