So to show this is a 2 manifold with boundary you have to show that around each point there is a neighborhood that is either homeomorphic to $D^2$ or $D^2_+= \{(x,y)\in \mathbb{R} | \,\,\,\, y\geq 0, \,\,\,\, |(x,y)|<1 \}$.
Let $X$ be the described set $X / \sim$ the quotient and $\pi$ the quotient homomorphisim.
For $x \in \pi( \text{int} \, ( X )) = \text{int} \, (X)$ we are done, this set is homeomorphic to the disk. On $\text{int} \,(X)$, $\pi$ is a homeomorphism.
For $x \in \pi( (-10, 10) \times \{1\})$ consider $\pi((-10, 10) \times [1,-1))$. Similarly for the other side.
For $x \in \pi( \{10\} \times (-1,1) )$ it is more difficult. Here we have to somehow work with the twist. Let $f: [-10,-9) \cup (9,10] \times (-1,1) / \sim \,\, \to (-1,1)^2 $ be:
$$
f(x,y) = \left\{
\begin{array}{lr}
(x-10,y) & : x \in (9,10] \\
(x+10,-y) & : x \in [-10,-9)
\end{array}
\right.
$$
I claim that this is continuous and bijective. Pulling $f$ back to $X$, i.e. considering $f \circ \pi : X \to (-1,1)^2$, it is continuous (this is the universal property of quotients). And it is bijective as $f \circ \pi$ is 1 to 1 except for the points that are identified where is is 2 to 1. But those points are identified so $f$ is 1 to 1 and onto. $f$ is also an open map, any open set in $X / \sim$ is the union of the images of an open sets from $X$ and $f \circ \pi$ is clearly an open map.
Note that if you cut a disk out of your lens shape, the remaining annulus is embedded in $\mathbb{R}^3$ without any "twists". If this could be "glued" to itself to form a Möbius band in $\mathbb{R}^3$, then the (external) boundary of your lens-shape region would be mapped to the "soul" of the Möbius band; namely, the curve running along the middle of the Möbius band for one of the standard embeddings.
However, it is easy to check that if one cuts the Möbius band open along its "soul", one gets a strip with two twists as you make full circle around it, rather than the strip with no twists that would result if one removed a disk from the "lens". Thus such a construction is impossible. Note that the use of the term "soul" is consistent with the term used in Riemannian geometry for manifolds of nonnegative curvature.
Best Answer
Topologically the full-twist strip and the untwisted strip are identical. The only difference is the way they are immersed in $\Bbb R^3$. So you aren't going to get an intrinsic characterization that is like the quotient space one you mentioned. (To see this, consider how you make a full-twist strip: take the untwisted strip, cut it across, twist it, and then rejoin each point on the cut to exactly the same points it was attached to before the cut. Or imagine twisting the strip in $\Bbb R^4$ instead of in $\Bbb R^3$.)
Knot theory faces a similar problem. Two linked circles are topologically identical to two unlinked circles: they are both the disjoint union of two circles. (Compare these with the boundaries of your two kinds of strips.) But knot theory wants to distinguish not the spaces themselves but the way they are immersed in $\Bbb R^3$. One way this is done is to consider the complement of the two sets in $\Bbb R^3$. Say the full-twist strip is $F_2$ and the no-twist strip is $F_0$. Then consider the spaces $ \Bbb R^3-F_2$ and $\Bbb R^3-F_0$. These are topologically distinct. But to really carry this out may be difficult.