Clean approach: given a vector bundle $E \to M$ of rank $n$, consider the action of $GL_n$ on $E \setminus 0$ (which is a $\Bbb R^n \setminus 0$ bundle), and on $\mu_2$ by the action of the determinant. Then the "orientation double cover of $E$" is the double cover $\Lambda(E) \to M$, where $$\Lambda(E) = (E \setminus 0) \times_{GL_n} \mu_2.$$

That is, all you're saying is "Forget the bases, just remember the orientation", and there is a way to make this happen just by taking products and quotients - insert a factor for arbitrary orientations, and then quotient by the rule saying a basis is equivalent to the corresponding orientation

Approach that is better for your conceptual understanding:

A vector bundle is specified by the following data: an atlas $\mathcal U$ on $M$, and maps $\rho_{ij}: U_i \cap U_j \to GL_n$, with $\rho_{ii} = 1$ and $\rho_{ij} \rho_{jk} = \rho_{ik}$, possibly up to me having put these in the wrong order.

We imagine that over each $U \in \mathcal U$ we have fixed isomorphisms $E\big|_U \cong U \times \Bbb R^n$, and the functions $\rho_{ij}$ are mediating the difference between these fixed isomorphisms over each $U_i \cap U_j$

but $\mu_2$ isn't a vector space

are you responded to "a vector bundle is..." ?

I'm not done

ok sorry

no worries

This is usually called the cocycle description of vector bundles; this data is precisely what is called a Cech cocycle in $Z^1(M; \mathcal{GL}_n)$, where the latter is the sheaf of groups given by $\mathcal{GL}_n(U) = \text{Map}(U, GL_n)$; one can perfectly make sense of first cohomology with coefficients in sheaves of nonabelian groups, and what the above description says is $$H^1(M; \mathcal{GL}_n) \cong \text{Vect}_n(M).$$

that mathcal looks ugly but ok.

There's no real reason here to restrict to $GL_n$. Here is the general definition: a fiber bundle $E \to M$ with fiber $F$ and structure group $G \subset \text{Aut}(F)$ (automorphisms continuous or smooth or whatever is appropriate here) is equipped with local trivializations $E \big|_U \cong U \times F$ as above, so that the transition maps $U_i \cap U_j \to \text{Aut}(F)$ always factor through $G$

Then a $G$-valued cocycle as above gives you via the same recipe a bundle $F \to E \to M$, determined up to isomorphism by the corresponding cohomology class $H^1(M; \mathcal G)$ - aka, essentially uniquely determined by the cocycle

For you, $F = G = \Bbb Z/2$

How does $GL_n$ act on $\mu_2$? I'm a bit slow

$A \cdot x = |\det(A)| x$

oh...

it seems I was being sloppy above, I apologize

it is indeed not $(E \setminus 0) \times_{GL_n} \mu_2$ as I guess you were confused about

I should have taken that to be $\text{Fr}(E) \times_{GL_n} \mu_2$, where the first is the fiber bundle with $$\text{Fr}_x(E) = \text{Space of bases for } E_x.$$

That's a pretty big mistake, sorry

In the second description, here is what's going on

We have a group homomorphism $GL_n \to \mu_2$, $A \mapsto |\det A|$

Therefore, if we have a bundle described by an atlas with transition functions $\rho_{ij}: U_i \cap U_j \to GL_n$ above, we may replace these by $|\det \rho_{ij}|: U_i \cap U_j \to \mu_2$

Because $|\det A|$ is a group homomorphism, these new transition functions still satisfy the cocycle law $\rho_{ij} \rho_{jk} = \rho_{ik}$

you mean $\operatorname{sgn}(\det A)$?

sorry

I'm trying to follow

So Fr(E) is a bundle of M?

I guess it's just the GL(n)-bundle of M, if that even makes sense?

@LeakyNun oh... yes, another mistake :(

it's alright...

@LeakyNun Exactly, these are called "Principal G-bundles", and we took a vector bundle and spat out a principal GL(n)-bundle

It's easier to pass between different principal bundles by the formal operations above, so it is nice to pass to that first

In general, "Fiber bundle with structure group $G$" and "principal $G$-bundle" carry the exact same data; you can go back and forth from one to the other (there's an eq. of categories)

The transition function stuff is mainly how I think about this in practice though... maybe in the last year I have converged towards the cleaner language but that's mostly because I haven't had to compute much lately or think in serious detail

and what does $\times_{GL_n}$ mean?

oh, sorry, that one I didn't realize I was being sloppy about

if $X$ has a right $G$ action and $Y$ has a left $G$-action, the "balanced product" $X \times_G Y := (X \times Y)/G$, with the equivalence relation precisely being $(xg, y) \sim (x, gy)$

the function for us above: replace the $GL_n$ fiber with a $\mu_2$ fiber, as $GL_n \times_{GL_n} X = X$