Boundary under transformation of a closed curve from $R^2\to R^3$ 
Consider some mapping $\phi: R_{uv} \to S\subset \mathbb{R}^3$ where
  $R_{uv}\subset \mathbb{R}^2$ and such that it is a simply connected
  region. We call the boundary of the surface (which we assume to be a
  regular closed curve) created by this mapping $\partial S=\phi(\partial R_{uv})$.
We only say that $S$ is a simple closed surface iff as a point moves along
  $\partial R_{uv}$ once, its image moves along $\partial S$ twice and in
  opposite directions.

Why (and how) does this point move twice through the boundary of the region? I'm not sure I understand the motivation behind this definition; could any one help with the intuition?
 A: An example of the phenomenon you describe could be the following: Let $R_{uv}$ be the rectangle
$$R_{uv}:=\{(u,v)\>|\>-\pi\leq u\leq \pi,\ -{\pi\over2}\leq v\leq{\pi\over2}\}\ ,$$
and consider the map
$$\phi:\quad R_{uv}\to{\mathbb R}^3,\qquad (u,v)\mapsto\left\{\eqalign{x(u,v)&:=\cos u\cos v,\cr 
y(u,v)&:=\sin u\cos v,\cr z(u,v)&:=\sin v.\cr}\right.$$
Then the two edges $v=\pm{\pi\over2}$ of $R_{uv}$ map to the two points $(0,0,\pm1)$, and the two edges $u=\pm\pi$ (directed upwards when $u=\pi$ and  downwards when $u=-\pi$) of $R_{uv}$ map on the same meridional arc of $S^2\subset{\mathbb R}^3$, but with opposite orientations. All in all you obtain a representation of $S^2$ of the required kind.
A: As I was out walking, I answered my own question.
Our mapping $\phi$ must divide the surface it is mapping onto in two sections (asymmetrical or otherwise---that is, in some sense, in order to 'unfold' it), each of which has a specific orientation. As such an orientation must have a smooth change, and must change at the boundary, we receive the consequence that the mapping causes our regular closed curve from our region $\partial R_{uv}$ to map once to $\partial S$ for one orientation and once to $-\partial S$ for the latter orientation (i.e. once 'forwards' and once 'backwards' through the curve), giving us the definition.
It's an ingenious definition, but not quite obvious for a first-read.
