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.
Yes, you get a deformation retract. More formally...
Suppose $M$ is a topological manifold, $N\subseteq \partial M \subseteq M$ is a component of $\partial M$ and that the inclusion $N\rightarrow M$ is a homotopy equivalence. Then $M$ deformation retracts to $N$.
Here's the idea of the proof:
From Hatcher Algebraic Topology, Corollary 0.20 (pg. 16 in my copy), it's enough to show that $(M,N)$ has the homotopy extension property.
Now, boundary components in topological manifolds are known to have collar neighborhoods (see Morton Brown, "Locally flat imbeddings of topological manifolds", Annals of Mathematics, Vol. 75 (1962), p. 331-341.) This is more well-known in the smooth category, where it is also much easier to prove.
So, some closed neighborhood of $N$ in $M$ is of the form $N\times [0,1] = N\times I$, where we are identifying $N$ with $N\times \{0\}\subseteq M$..
Now suppose $f:M\rightarrow X$ is a continuous function (where $X$ is any topological space). Suppose $F:N\times I\rightarrow X$ has the property that $F(n,1) = f(n)$. We wish to find a continuous function $G:M\times I\rightarrow X$ satisfying two conditions. First, for any $n\in N$, $t\in I$, that $G(n,t) = F(n,t)$. Second, for any $m\in M$, $G(m,1) = f(m)$.
To that end, set $G(m,t) = \begin{cases} f(m) & m\notin N\times [0,1]\\ F(n, \max\{s,t\}) & m=(n,s)\in N\times [0,1].\end{cases}$
To check the first condition, note that $n\in N\cong N\times \{0\}$ means the corresonding $s$ is $s= 0$. Since $t\in[0,1]$, $\max\{s,t\} = \max\{0,t\}= t$, so $G(n,t) = F(n,t).$
To check the second condition, we have $G(m,t) = f(m)$ if $m\notin N\times [0,1]$, regardless of the value of $t$. On the other hand, if $m = (n,s)\in N\times [0,1]$, then because we are taking $t=1$, we have $\max\{s,t\} = t$. So $G(m,1) = F(m,1) = f(m)$.
Finally, we must verify that $G$ is actually a continuous function. Since $f,F,$ and $\max$ are all continuous, we see that $G$ is a continuous funciton ... as long as it is a function. That is, we still need to verify that $G$ is well defined at the "transition" points where $s = 1$.
But, if $m\in N\times [0,1]$ is of the form $m = (n,1)$, then $s=1$, so $\max\{s,t\} = s=1$, so $G(m) = F(n, \max\{s,t\}) = F(n,1) = f(n).$
Best Answer
The notion of boundary that you are looking for comes from the definition of topological manifolds with boundary. As opposed to a regular manifold $X$, a manifold with boundary has the property that each point in $X$ has an open neighborhood which is homeomorphic to an open set in the euclidean half space $\mathbb{R}_+^n=\{(x_1,\dots,x_n)\in\mathbb{R^n}:x_n\ge0\}$. Thus we then define $\partial X$ to be the points which when mapped to $\mathbb{R}_+^n$ have $x_n=0$.
This definition has the benefit that an embedding of $X$ into some other space does not change $\partial X$. Thus $\partial D^2=S^1$ irregardless of whether you view it as living in $\mathbb{R}^2$ or in $\mathbb{R}^3$.