Let $\{M\}$ be the one-sided Moebius strip and $\{MM\}$ the corresponding two-sheeted Moebius strip.
Let us assume $\{MM\}$ to be the doubling of $\{M\}$; hence it is orientable.
Then, $\{M\}$ is orientable, too?
I assume the definition of doubling from:
A Texbook of Topology, Seifert and Threlfall:
~24*: Moebius band is the simplest example of a non-orientable pseudomanifold with boundary.
36: We define doubling for a pure complex: we take a homeomorphic copy...and we identify points of the boundary...with points of the boundary... A pseudomanifold with boundary is a bounded pure complex whose doubling is a closed pseudomanifold. The pseudomanifold with boundary is orientable if and only if its doubling is orientable.https://en.wikipedia.org/wiki/Glossary_of_differential_geometry_and_topology:
Doubling, given a manifold M with boundary, doubling is taking two copies of M and identifying their boundaries. As the result we get a manifold without boundary.The Geometry of Physics by T. Frankel:
16.11: Each non-orientable manifold has a 2-sheeted orientable covering manifold...
UPDATE To answer to @Christian Blatter (I'm new to this site, sorry I'm not yet allowed to answer...):
I admired your symbolism. It is only a symbolism, but how useful it is! By it, shortly: $D_{\rm ST}(M)$ a closed manifold; ${\rm DC}(M)$ a manifold with boundary. Surely I used $D_{\rm ST}(M)$ in a cavalier manner.
I add an image link here to describe my case study: Moebius Strip Composition
A 2D Intrinsic Observer, with left, front, right, aft, no top, no bottom.
Left hand, one-sided, sheet #1 Moebius strip ${\rm M1}$, twist 0->Π rad, with closed boundary of lenght ~2(L+l).
In series to it, in the same cover: the left hand, one-sided, sheet #2 Moebius strip ${\rm M2}$, twist Π->2Π rad, with closed boundary of length ~2(L+l).
The length of the closed Midline of ${\rm M1}$ being that ${\rm L}$ of the generating one-sided plain rectangular (L*l) Strip ${\rm S}$. The boundary of ${\rm M1}$ is equivalent to the boundary of ${\rm S}$. The same for ${\rm M2}$.
The above pair yielding the (assumed transparent), one-sided, two-sheeted, oriented manifold ${\rm MM}$, total twist 0->2 Π rad; the former superposed boundaries reduced to an inner curve.
I must analyze ${\rm DC}(M1)$ and $D_{\rm ST}(M1)$ in ${\rm MM}$.
The doubling ${\rm ST}(M)$ of a manifold with boundary $M$ according to the definition in Seifert-Threlfal is completely different from a double cover ${\rm DC}(M)$ of $M$. Note that $D_{\rm ST}(M)$ is a closed manifold whereas the double cover ${\rm DC}(M)$ doubles the boundary as well, hence is again a manifold with boundary.
You get the double cover of a Moebius strip $M$ if you first manufacture a model of $M$ from paper. Then each point $p\in M$ is a puncture of the paper, and is visible on both sides. You obtain the double cover ${\rm DC}(M)$ of $M$ by viewing the two aspects of the puncture on the two sides of the paper as two different copies of $p$. By walking from $p\in M$ once around $M$ you arrive at the same puncture, but you have to walk twice around $M$ to arrive at the same point in ${\rm DC}(M)$.