Why orientation of plane region boundary implies surface boundary? For parametrized surfaces $\Phi:D\to \mathbb R^3,$ we assume that $\Phi$ is one to one on the whole of $D$. Then the image of $D$ is $S$ and we define the image of $\partial D$ (the boundary of $D$) is  $\partial S$ (But I'm sure I'm missing something, eg., we know boundary of sphere (as surface $\mathbb R^3$) is empty).

Question: (1)If we assume that $\partial D$ is oriented as in Green's theorem, Can we get an orientation for $\partial S$ and $S$ itself is oriented so $\Phi_u \times \Phi_v$ is in the direction of the normal.  (2) Is the above definition of boundary  of surface  is correct in $\mathbb R^3$? If so, then I think  then $\partial S \neq \emptyset$


If $D$ is a  plane  region, with boundary curve $C,$ the positive orientation of $C$ is given by  the vector $k\times v_{out},$ where $v_{out}$ is the normal vector field pointing outward along curve.
As you walk along $C$ in the direction of positive orientation, the region $D$ is on your left.
Edit: Assume that $D$ is  plane region of elementary type (as in Greens theorem)
 A: I think what's going on is that you are thinking about a somewhat restricted version of Stokes Theorem, which I'll refer to as the "singly parameterized Stokes Theorem", in which one assumes that the surface $S$ is restricted to be a parameterized surface, with parameterization $\Phi : D \to S$ as stated in your question. In that case all I can say is that the answers to your two questions are "Yes" and that $\partial S$ is indeed nonempty, for the reasons that you suggest.
However, there is a more general version of Stokes Theorem where $S$ is not assumed to be "singly parameterized", and where $\partial S$ is allowed to be empty. In this more general situation there is no guarantee that $S$ is orientable (the Möbius band is a nonorientable, for example), and therefore the more general version has an extra hypothesis requiring that $S$ be orientable, and another hypothesis that $S$ be compact. The proof of this general version of Stokes Theorem does make use of the multiple coordinate systems needed to cover an arbitrary oriented surface $S$.
For example this more general version of Stokes Theorem does indeed apply to the case that $S \subset \mathbb R^3$ is the unit 2-sphere (which means the boundary of the unit 3-ball in $\mathbb R^3$). Of course we know intuitively that the unit 2-sphere is not a "singly parameterized" surface. For example spherical coordinates do not apply to the north or south pole, so we have to add a "north polar" coordinate system and a "south polar" coordinate system, which gives a covering of the unit 2-sphere by three coordinate systems. You can actually do better and cover the unit 2-sphere with only two coordinate systems. But there does not exist any single coordinate system that covers the whole 2-sphere in a one-to-one fashion.
It is difficult to explain the details of coordinate systems, orientations, and boundaries without some more advanced knowledge of manifolds, which one eventually learns about in a course on differential geometry or differential topology.
