I take it that $S^1$, for you, is the unit interval with an identification of the two ends, so that $0 \sim 1$.
The intersection of $U_1$ and $U_2$ is then messy, because it has two components, namely $0 < x < 1/2$ and $1/2 < x < 1$. You want to define $g_{12}$ on both parts, and I suggests that you define it on the first part by
$$
g_{12} (x)(y) = y
$$
while on $1/2 < x < 1$ you define it by
$$
g_{12} (x)(y) = 1-y.
$$
The result should THEN be a mobius band. What you've got a this point is a cylinder (as you've verified).
I'm going to slightly edit your definition, and say that I want
$g_{12}(x)(y) = -y$ for $1/2 < x < 1$ (and similarly for the Mobius band), because this leaves a "centerline" at $y = 0$, while in your definition, the centerline is at $y = 1/2$; if you were defining a finite mobius band, using $[0, 1] \times [0, 1]$, that might make sense (although I might argue for $[0, 1] \times [-1, 1]$ instead!). But since you're using all of $\mathbb R$ as a fiber, it'd be nice if the transition maps were, fiberwise, vector-space isomorphisms, which requires sending $0$ to $0$.
Now my space $X$ consists of equivalence classes that look like one of four things:
$$
L_{a, s} = \{(a, s, 1), (a, s, 2) \}
$$
(where I've added an integer "identifier" to each patch), and where $s$ is any real number, or
$$
R_{a, s} = \{(a, s, 1), (a, -s, 2) \}
$$
The first is for $0 < a < 1/2$, and the second for $ 1/2 < a < 1$. Then there are two other kinds of equivalence classes:
$$
Z_s = \{(0, s, 1), (1, -s, 2) \}
$$
and
$$
H_s = \{ (\frac{1}{2}, s, 1), (\frac{1}{2}, -s, 2) \}.
$$
(The "Z" is mnemonic for "zero or one" and the "H" is for "halfway".)
In the Mobius band $M$, there are two kinds of equivalence classes:
$$
U_{a, s} = \{(a, s)\}
$$
(the "U" is for "usual") and
$$
P_{s} = \{(0, s), (1, -s)\}
$$
where the "P" is for "excePtional". :)
Now all I have to do is tell you how my homeomorphism $F$ will send each class in $X$ to a class in $M$. In the following, each line of the definition holds for all $s \in \mathbb R$. So:
\begin{align}
F(L_{a,s}) &= U_{a, s} & 0 < a < \frac{1}{2} \\
F(R_{a,s}) &= U_{a, -s} & \frac{1}{2} < a < 1 \\
F(Z_{s}) &= P_s \\
F(H_{s}) &= U_{ \frac{1}{2}, s}.
\end{align}
It's not hard to see that $F$ is bijective, and I leave that to you; the only question is continuity.
Continuity except at $a = 0, \frac{1}{2}, 1$ seems pretty clear, so I won't discuss that. Let's check continuity at $a = \frac{1}{2}$. To do so, I'm going to look at a "vertical" curve in $X$ and a "horizontal" one, and check that each maps to a nice continuous curve in $M$, That's not really a proof, but I'm hoping it'll suffice to convince you.
The vertical curve is
$$
\gamma: I \to X : t \mapsto H_t.
$$
Under the map $F$, this becomes
$$
\alpha = F\circ \gamma : I \to M : t \mapsto F(H_t) = U_{ \frac{1}{2}, t}
$$
which looks perfectly nice (as it should: $F$ is a vector-space isomorphism on the fiber!).
What about a "horizontal" curve in $X$? Let's look at one at "height" $1/3$:
$$
\beta: [\frac{1}{4}, \frac{3}{4}] \mapsto X : t \mapsto \begin{cases}
L_{t, \frac{1}{3}} & t < \frac{1}{2} \\
H_\frac{1}{3} & t = \frac{1}{2} \\
R_{t, \frac{1}{3}} & t > \frac{1}{2}
\end{cases}.
$$
It's pretty clear that $\beta$ is a continuous curve, as is $F\circ \beta$. So that part works out OK as well.
Finally, there's continuity at $0$. That's the tricky one. Once again, I'll use $\gamma$ and $\beta$ to denote the vertical and horizontal curves. In fact, I won't even bother with $\gamma$ -- you can give that a shot yourself. But $\beta$ is more interesting. I'll define it for $-1 < t < 1$, to make life a little easier for myself:
$$
\beta: [-1, 1] \to X: t \mapsto
\begin{cases}
Z_\frac{1}{3} & t = 0 \\
L_{\frac{t}{2}, \frac{1}{3}} & t > 0\\
R_{1+\frac{t}{2}, -\frac{1}{3}} & t < 0
\end{cases}
$$
If you look at $F\circ \beta$, you'll see that it's a curve that runs along at height $1/3$ in the mobius band to the right of $x = 0$, and at height $-1/3$ to the left of $x = 1$, which makes it continuous in the Mobius band. Of course, you also need to check that it's continuous in $X$, but I think that's not TOO difficult.
I hope (a) that this explicit description is of some help to you, and (b) that you never have to do such a thing explicitly again. :)
Best Answer
Consider $[x,y]\in r(\{0,1\}\times(0,1))$. Note that $[x,y]=\{(0,t),(1,1-t)\}$ for some $t\in(0,1)$, as an equivalence class. Pick $0<r<\min(t,1-t)$ and consider two half open disc:
$$D_1=\big\{v\in I^2\ \big|\ \lVert v-(0,t)\rVert < r\big\}$$ $$D_2=\big\{v\in I^2\ \big|\ \lVert v-(1,1-t)\rVert < r\big\}$$
These are two disjoint and open subsets of $I^2$. They are not standard disks however we can define a continuous function
$$h:D_1\cup D_2\to \mathbb{R}^2$$ $$h(x,y)=\begin{cases} (x,y) &\text{if }x\leq 1/2 \\ (x-1, y-1) &\text{otherwise} \end{cases}$$
and note that the image of $h$ is precisely the open ball $B$ in $\mathbb{R}^2$ around $(0,t)$ of radius $r$. Of course $h$ is not a homeomorphism, because it is not injective, however it induces an injective map $H:(D_1\cup D_2)/\sim\to B$ on the quotient space, and thus $H$ is a homeomorphism. Note that $(D_1\cup D_2)/\sim$ is an open neighbourhood of $[x,y]$.
In particular $[x,y]$ has an open neighbourhood homeomorphic to $\mathbb{R}^2$ and thus it cannot be a boundary point.
For that you can utilize the fact that if $v\in M$ is such that any open neighbourhood $U$ of $v$ satisfies $U\cap\partial M\neq \emptyset$ then $v\in\partial M$. This is a simple consequence of the fact that $\partial M$ is a closed subset of $M$. Which follows (as earlier) from the observation that $x\in M\backslash\partial M$ if and only if $x$ has an open neighbourhood homeomorphic to $\mathbb{R}^n$.