As far as my understanding goes, if one has a closed (compact) surface and "cuts out" a disk, then the resulting object has a non-empty/non-trivial boundary.
It can be shown that any non-orientable surface without boundary cannot be embedded into $\mathbb{R}^3$ (e.g. A non orientable closed surface cannot be embedded into $\mathbb{R}^3$, https://mathoverflow.net/questions/18987/why-cant-the-klein-bottle-embed-in-mathbbr3).
However, if we allow surfaces with boundary, then we can find a non-orientable surface which does embed into $\mathbb{R}^3$, the Möbius strip. As I understand, it can be formed by cutting out a disk from the real projective plane $\mathbb{RP}^2$. (Which corresponds to the fact that the boundary of both the disk and of the Möbius strip is a circle.)
Thus, it seems like the pathology which prevents $\mathbb{RP}^2$ from being embedded into $\mathbb{R}^3$ is the disk we cut out to get the Möbius strip (I think this disk is sometimes called the cross-cap, see here).
Is there a way to make rigorous the notion that only "certain types of twisting" are the only things which prevent a surface from being embedded into $\mathbb{R}^3$?
An example of such a twist is the twist necessary to glue a disk onto the Möbius strip.
In other words, why does introducing boundaries to non-orientable surfaces (by cutting out a disk) suddenly make them embeddable?
The concept that a manifold with boundary could be less pathological than similar manifolds (i.e. of the same dimension and orientability) that do not have boundary is counter-intuitive for me, and want to understand what I am failing to take into account.
The fact that the Möbius strip has boundary clearly gives it interesting topological properties which non-orientable surfaces without boundary do not have, see for example this video on YouTube, however I don't have the impression that the video demonstrates how the boundary of the Möbius strip makes it embeddable into $\mathbb{R}^3$.
