I think I (finally) understood your question. You know how to obtain surfaces (such as Moebius bands) by identifying sides of polygons. What you are asking is for a rigorous description of a "cut-and-paste operation on the Euclidean 3-space $R^3$ which produces the doubly twisted Moebius band in $R^3$."
I will write down an answer but it might take you awhile to understand it. (For the record: This is not the best way to get "doubly twisted Moebius bands": A better way is to used a "framed unknot": Say, a round circle $C$ equipped with a normal vector field that twists once around $C$.)
Here is one way to make this work. Start with $R^3$ and a "vertical round annulus" $A$ contained the $xz$-plane. The boundary of $A$ is the union of two round circles $C_1, C_2$, which I will assume to be concentric with the common center at the origin $(0,0,0)$ and respective radii 1 and 2. Then $C_1$ intersects the $xy$-plane in two points $p_1=(-1,0,0), q_1=(1,0,0)$, while $C_2$ intersects the $xy$-plane also in two points $p_1=(-2,0,0), q_1=(2,0,0)$.
Now, draw a round circle $\alpha$ in the $xy$-plane which separates $p_1,q_1$ from $p_2,q_2$. The precise location of this circle is irrelevant.
Consider the surface $P$ equal to the $xy$-plane and let a homeomorphism $f: P\to P$ be the Dehn twist along the loop $\alpha$. (For the record, there is a left and right Dehn twists, it does not matter for our purpose which one do we use.) This homeomorphism will be identity outside of a small neighborhood of the loop $\alpha$. In particular, it is the identity on the intersection $A\cap P$.
Let $P_\pm$ denote the lower and upper half-spaces in $R^3$ bounded by the plane $P$. I will define a new topological space $X$ obtained by gluing $P_-$ to $P_+$ using the homeomorphism $f$ of their boundaries. The annulus $A$ still sits inside $X$ (since $f$ was the identity at $A\cap P$). One can show that $X$ is homeomorphic to $R^3$ (for instance, because $f: P\to P$ is isotopic to the identity map). However, under the homeomorphic identification $X\to R^3$ the annulus $A$ maps to a "once overtwisted annulus" (doubly twisted Moebius band that you are interested in). Proving that this is true will take some effort and I will not do it. I note only that if instead of gluing using $f$ you perform gluing using its $n$th power, you obtain an $n$-times overtwisted annulus in $R^3$. One can also use a "half-twist" $g$ instead of $f$ ($g$ has the property that $g\circ g=f$: It is a 180 degree rotation inside the circle $\alpha$ in $P$ and swaps $p_1, q_1$ and fixes the midpoint of $p_1q_1$). If you glue $P_+, P_-$ using $g$ you obtain a Moebius band embedded in $R^3$ in a standard way. You can also replace $g$ with its odd powers. This will result in overtwisted Moebius bands in $R^3$.
In order to understand what is going on, I suggest you draw a picture.