Orientation Sheaf and Double Cover The orientation sheaf of an $n$-manifold $M$ is $\mathcal{O}_n=Sheaf(U\mapsto H_n(M,M-U;\mathbb{Z}))$, with stalks given by $(\mathcal{O}_n)_x = lim H_n(M,M-U)=H_n(M,M-x)=\mathbb{Z}$ (the limit is over neighborhoods $U$ containing $x\in M$).
Suppose $M$ is orientable and closed, so that $H_n(M)=\mathbb{Z}$. Then $\mathcal{O}_n=M\times\mathbb{Z}$ is a constant sheaf.
Suppose now that $M$ is nonorientable.  Then we can take a neighborhood $N$ of each point $x\in M$ so that $\mathcal{O}_n|_N$ is constant.  Thus $\mathcal{O}_n$ is a locally constant sheaf.
The orientable double cover of $M$ is $\tilde{M}=\lbrace(x,\mu_x)\rbrace$ where $\mu_x$ is a local orientation at $x\in M$ (a choice of generator of the local homology).   If $M$ is nonorientable then $\tilde{M}$ is connected, and if $M$ is orientable then $\tilde{M}\approx M\times\mathbb{Z}_2$.  In either case, there is an embedding $\tilde{M}\hookrightarrow \mathcal{O}_n$.  
Is there anything else I can say about this enlargement from $\tilde{M}$ to $\mathcal{O}_n$ ?  Does one encode more information than the other?
If we instead use $\mathbb{R}$-coefficients, I believe (but can easily be wrong) that $\tilde{M}$ is a deformation retract of $\mathcal{O}_n$ for nonorientable $M$, and is a deformation retract of $\mathcal{O}_n-\Gamma^0$ for orientable $M$ (where $\Gamma^0$ is the zero section of this real line bundle).  In other words, $\tilde{M}$ and $\mathcal{O}_n$ are the same up to homotopy (over $\mathbb{R}$-coefficients), and one does not encode more information than the other.
 A: An elaboration.
$\mathcal O_n$ is a bundle over $M$ with fiber $\mathbb Z$.  There is an action of the integers on it, because the integers act on homology.  Earlier I said this was a principal bundle, I was too tired!  The action is of course not free.  In particular, this bundle $\mathcal O_n \to M$ has a section. 
Given a manifold $M$, its orientation cover is a $\mathbb Z_2$-principal bundle.  The action of $\mathbb Z_2$ is given by reversal of orientations. 
So the product $\tilde M \times \mathbb Z$ has an action of $\mathbb Z_2$ given by 
$$t.(x,n) = (t.x, (-1)^tn)$$
where $t.x$ is the action of $\mathbb Z_2$ on $\tilde M$. 
$\tilde M \times_{\mathbb Z_2} \mathbb Z$ is the quotient of $\tilde M \times \mathbb Z$ by the action of $\mathbb Z_2$.  By design, it is a $\mathbb Z$-principal bundle -- you can project onto $\tilde M / \mathbb Z_2 \equiv M$ -- and by design it is isomorphic to $\mathcal O_n$.  The map $\tilde M \times_{\mathbb Z_2} \mathbb Z \to \mathcal O_n$ is induced by your inclusion $\tilde M \to \mathcal O_n$. 
