A continuously differentiable map ${\bf f}: \quad\Omega\to{\mathbb R}^3,\quad (u,v)\mapsto{\bf x}(u,v)=\bigl(x_1(u,v),x_2(u,v),x_3(u,v)\bigr)$ defined on an open set $\Omega\subset{\mathbb R}^2$ is called an immersion, if for all $(u,v)\in\Omega$ the differential $d{\bf f}(u,v)$ has rank $2$, i.e., if ${\bf x}_u\times{\bf x}_v\ne{\bf 0}$ for all $(u,v)$.
When this "technical condition" is fulfilled then ${\bf f}$ maps sufficiently small pieces of $\Omega$ bijectively onto small pieces of a (large) smooth surface $S:={\bf f}(\Omega)$. But globally the map ${\bf f}$ need not be one-one: There might be self-intersections, or some parts of $S$ are covered several times.
Such is the case in your example: One computes ${\bf g}_u\times{\bf g}_v=(a+b\cos u)\bigl(-b\cos u\cos v, -b \cos\sin v,-b\sin u\bigr)\ ,$ which is easily seen to be $\ne{\bf 0}$ for all $(u,v)\in{\mathbb R}^2$. Therefore this ${\bf g}$ is indeed an immersion. But the function ${\bf g}$ is doubly periodic; whence any two points $(u,v)$ differing by $(2k\pi,2\ell\pi)$ are mapped to the same point of $S$. Nevertheless it makes sense to leave the map ${\bf g}$ as it stands, because restricting $u$, $v$ to a period square introduces artificial seams.
It is another matter when you want to compute the area of $S$. In this case you have to make sure that $S$ is covered exactly once by the parametrization (up to said seams, which form a set of measure zero).