Thomas' Calculus puts forward the following statement:
The requirement that $\vec r(u, v)$ be one-to-one on the interior of a region $\mathrm R$ ensures that the surface does not cross itself.
So far I have not come across any reason behind why this requirement is necessary. I, however, have given this statement some thought. The following is what I could reason. Corrections are welcome.
Let us calculate the area of a hemisphere( radius one unit ) with the parametrization, $$\vec r(u, v) = \sin(v) \cos(2u)\hat i +\sin(v) \sin(2u)\hat j + \cos(v) \hat k, 0\leq v \leq \pi; 0\leq u \leq 2\pi$$
The surface shown above overlaps on itself once. The area can be calculated using, $$\iint\limits_{\mathrm R} |\vec{r_u} \times \vec{r_v}| \mathrm du\mathrm dv$$
We, using our parametrization, find that the area is $4\pi$( units squared ). This is the area of a sphere and the not the hemisphere whose area we intend to calculate. Had we restricted $u$ to $\pi$, we would have had a one-to-one parametrization yielding the correct area.
You have the right idea. If your parametrization is not one-to-one, you integrate over the same 'patch' of surface multiple times, resulting in the wrong area.
If for some reason you want to use a non-injective function $r$, you can modify the area formula to read
$$A = \iint_R |r_u \times r_v| \frac{1}{n(u,v)}\;dudv$$ where $n(u,v)$ is the number of pre-images of $r(u,v)$ in $R$ (that is, the number of points $(u', v')$ such that $r(u',v') = r(u,v)$). This formula will work provided every point on the surface has a finite number of pre-images. For example, under the parametrization you gave for the hemisphere, almost every * point has $2$ pre-images, so $n(u,v) = 2$ (almost everywhere * ). Using the modified formula, you get the correct surface area: $$\iint_R |r_u \times r_v|\frac{1}{n(u,v)}\;dudv = \frac{1}{2} \iint_R |r_u \times r_v|\;dudv = 2\pi.$$
*: The points on the half-meridian $r(0,v)$ have $3$ pre-images, since $r(0,v) = r(\pi,v) = r(2\pi,v)$. However, the half-meridian has $0$ area, so it doesn't matter that our formula is wrong here.