I'm trying to better understand surface integrals involving vector fields, and I require that the surface be orientable. As I understand it, a non-orientable surface is one that does not have a unique choice of normal vector.
Although I'm certain that there are tons of subtlety to the definition, is the following a sufficient (if not necessary) condition:
Let $f(x,y,z) = 0$ be a surface. Then $f$ is orientable over some domain $D$ if $\nabla f$ is a continuous function everywhere in $D$.
Suppose that you have a surface $S$ in $R^3$ given by $S=\{{\mathbf x}: f({\mathbf x})=0\}$, where $f\in C^1$ and has nonvanishing gradient on $S$. Then $S$ is smooth (by the implicit function theorem) and oriented since its normal vector field can be given by $\nabla f$.
Without the nonvanishing assumption on $\nabla f$, $S$ may fail to be even a topological surface, can fail to be a smooth surface and can even be a smooth nonorientable surface.
If $S$ is the graph of a continuous function $g: D\to R$, where $D\subset R^2$ is an open subset then $S$ is always orientable since it is homeomorphic to $D$. If, in addition, $g\in C^1$ then $S$ satisfies the assumption of item 1 and hence, admits a normal vector field.