Properties of Orientation Covering: Lee's Construction

I think it is only a notational issue.

In my opinion it is misleading to write $(p,\mathcal O_p)$ for the points of $\hat M$ because that suggests that we have a function assigning to $p \in M$ an orientation $\mathcal O_p$ of $T_pM$. I would neutrally write $(p, \mathcal O)$, where $\mathcal O$ is one of the two orientations of $T_pM$.

Via the map $\hat \pi : \hat M \to M, \hat \pi( p, \mathcal O) = p$, we obtain a topology on $\hat M$ such that $\hat \pi$ becomes a covering projection with two sheets. This is due to the fact that $M$ is locally orientable. That is, if we take an orientation $\Omega_U = (\omega_p)_{p \in U}$ on some open $U \subset M$ (such $\Omega_U$ always exists if $U$ carries a chart), then the two sheets over $U$ are given by $U_\pm = U_\pm(\Omega_U) = \{ (p,\pm \omega_p) \mid p \in U \}$.

Moreover, we get a smooth atlas on $\hat M$ such that $\hat \pi$ is smooth.

Now give $T_{(p,\mathcal O)} \hat M$ the unique orientation $\omega_{(p,\mathcal O)}$ such that such $T_{(p,\mathcal O)} \hat \pi : (T_{(p,\mathcal O)} \hat M, \omega_{(p,\mathcal O)}) \to (T_p M, \mathcal O)$ is orientation preserving. Intuitively, the orientation on $T_{(p,\mathcal O)} \hat M$ is just the given orientation $\mathcal O$ at the point $(p,\mathcal O)$. This procedure has nothing to do with any local (or global) orientation on $M$.

Then one can show that $\hat \Omega = (\omega_{(p,\mathcal O)})_{(p,\mathcal O)}$ is an orientation on $\hat M$. It is defined canonically without making any choices.

Edited:

Let $\mathfrak A$ be the smooth subatlas on $M$ consisting of all charts $\phi : U \to V$ with connected $U$. Connectedness assures that $U$ has exactly two orientations. Let us call an open $U \subset M$ nice if it is occurs as the domain of a chart in $\mathfrak A$. For nice $U$, the set $(\hat \pi)^{-1}(U)$ can be decomposed uniquely as the disjoint union of two subsets $U_\pm = U_\pm(\Omega_U)$ as above. This decomposition does not depend on the choice of $\Omega_U$. In fact, we have $U_\pm(-\Omega_U) = U_\mp(\Omega_U)$ which yields the same two sets. Only the indexing by $\pm$ depends on $\Omega_U$. By construction of $U_\pm$, $T_{(p,\mathcal O)} \hat \pi : (T_{(p,\mathcal O)} \hat M, \omega_{(p,\mathcal O)}) \to (T_p M, \omega_p)$ is orientation preserving for $(p,\mathcal O) \in U_+(\Omega_U)$ and orientation reversing for $(p,\mathcal O) \in U_-(\Omega_U)$.

The set of all $U_+$ and all $U_-$ forms the basis of a topology on $\hat M$ and it is easy to see that $\hat \pi$ becomes a covering projection with two sheets. An atlas $\hat{\mathfrak A}$ on $\hat M$ is given by the maps $\phi_\pm = \phi \circ \hat \pi \mid_{U_\pm} : U_\pm \to V$. This atlas is clearly smooth. That is, $\hat M$ has been given the structure of a smooth manifold such that $\hat \pi$ becomes smooth.

The family $\hat \Omega = (\omega_{(p,\mathcal O)})_{(p,\mathcal O)}$ is by construction locally an orientation. In fact, it is an orientation on each of the sets $U_\pm(\Omega_U)$. Thus it is an orientation on $\hat M$. This is true for any smooth manifold $N$ and any family of orientations $(\omega_q)_{q \in N}$ of the tangent spaces $T_qN$ which is locally an orientation. Observe that the existence of such a (global) family does not follow from the obvious fact that each smooth manifold $N$ is locally orientable. Indeed, there is no reason to assume that we can paste a suitable set of local orientations to a global family of orientations $(\omega_q)_{q \in N}$ which restricts to all given local orientations.

Note also that $\hat{\mathfrak A}$ is not an oriented atlas. It only contains oriented atlases.