Getting a true answer to your question requires some familiarity with differential topology (or algebraic topology), which I assume in what follows. My favorite reference is do Carmo's "Riemannian Geometry".
The definitions below are (intentionally) formal (you had enough informal input from other answers). To gain an intuition, one should work out specific examples of surfaces in $R^3$.
In order to keep things simple, I will assume in what follows that $S$ is a smooth hypersurface in an $n$-dimensional Riemannian manifold $(M,g)$, i.e., $S$ is a smooth submanifold in $M$ and $S$ has dimension $n-1$. (The assumptions of smoothness and presence of a Riemannian metric, simplify definition and proofs but can be avoided by doing a bit more work. To avoid using Riemannian metric I would have to define the normal bundle of $S$ in $M$. To eliminate smoothness assumptions, I would have to invoke some heavy machinery from the theory of topological manifolds.) For concreteness, you can consider the case when $M=R^n$ with the standard metric.
Definition. A unit normal vector to $S$ at a point $x\in S$ is a unit vector in $T_x(M)$ (the tangent space of $M$ at $x$) which is orthogonal to $T_x(S)$ (the tangent space of $S$ at $x$).
Definition. A unit normal vector field to $S$ is a vector field along $S$ consisting of unit normal vectors, i.e., a smooth map $\nu: S\to TM$ (the tangent bundle of $M$) which sends each $x\in S$ to a unit normal vector $\nu_x$ to $S$ at the point $x$.
Definition. A coorientation of $S$ is a choice of a unit normal vector field $\nu$ to $S$. A submanifold $S\subset M$ is called coorientable if it admits a coorientation.
Clearly, if $\nu$ is a coorientation of $S$, then the vector field $-\nu$ is also a coorientation of $S$, and $-\nu\ne \nu$.
Exercise. Coorientability of $M$ is independent of the choice of a Riemannian metric on $M$: $S$ is coorientable if and only if there exists a continuous vector field $\xi$ along $S$ such that $\xi_x\notin T_xS$ for every $x\in S$. Hint: Use the orthogonal projection of $T_xM$ to the orthogonal complement of $T_xS$ in $T_xM$.
Note that coorientability is independent of orientability of $S$ and $M$. It is also independent of the number of components of $M\setminus S$. However:
Exercise. a. If both $M$ and $S$ are orientable, then $S$ is coorientable.
b. If $M$ and $S$ are connected and $M\setminus S$ is not connected, then $S$ is cooriented.
Definition. Suppose that $S$ is connected. Then $S$ is called 1-sided if it does not admit a coorientation. (Such $S$ is said to have one side.)
Lemma. Suppose that $S$ is connected and coorientable. Then $S$ admits exactly two coorientations.
Proof. Let $\nu, \mu$ be coorientations of $S$. I claim that either $\nu=\mu$ or $\mu=-\nu$. Consider a point $x\in S$. Then either $\nu_x=\mu_x$ or $\nu_x=-\mu_x$ since these are unit normal vectors to $S$ and $T_x M= T_x S\oplus {\mathbb R}$ is the orthogonal decomposition. Therefore, we obtain a partition $U\sqcup V$ of $S$ where
$$
U=\{x\in M: \nu_x=\mu_x\}, V=\{x\in M: \nu_x=-\mu_x\}.
$$
Both sets are closed in $S$ since $\nu_x, \mu_x$ are continuous vector fields. If both sets are nonempty then $S$ is not connected, which contradicts our assumption. Hence, either $\nu_x=\mu_x$ for every $x\in S$ or $\nu_x=-\mu_x$ for every $x\in S$. qed
Definition. A side of a connected hypersurface $S$ is a choice of coorientation of $S$.
Corollary. Every connected hypersurface either is 1-sided or has exactly two sides. In other words, the number of sides, $\sigma(S)$, of a connected hypersurface $S$ is either 1 or 2.
Suppose now that $S$ is not necessarily connected and
$$
S= \coprod_{j\in J} S_j
$$
is the decomposition of $S$ in its connected components. Then the number of sides of $S$ is defined as
$$
\sigma(S):= \sum_{j\in J} \sigma(S_j).
$$
For instance, if $S$ is the disjoint union of the Moebius band and an annulus in $R^3$, then $\sigma(S)=1+2=3$, that is, $S$ has three sides. If $M=RP^2\times S^1$ and $S=RP^2\times \{p\}\subset M$, then $\sigma(S)=2$, that is, $S$ has two sides.
This answers your question on the "number of sides" of a hypersurface. Note that if $S$ is not a hypersurface in $M$ then its "number of sides" is not defined.
][1]
(source: [oberlin.edu](http://www.oberlin.edu/math/faculty/bosch/smiley2006-yellow-red-dot-web.png))
