Understanding the orientable double cover 
Definition: if $M$ is a smooth manifold, define the orientable double cover of $M$ by:
$$\widetilde{M}:=\{(p, o_p)\mid p\in M, o_p\in\{\text{orientations on }T_pM\}\}$$
together with the function $\pi:\widetilde{M}\to M$ with $\pi((p,o_p))=p$.

There are three things I'm trying to understand about $\widetilde{M}$:

*

*What is its differentiable structure?


*Why is $\widetilde{M}$ orientable?


*Why is the connectedness of $\widetilde{M}$ equivalent to the non-orientability of $M$?
Here's where I'm at: first, for the topology of $\widetilde{M}$, one may define $\widetilde{U}\subset\widetilde{M}$ as open $\Leftrightarrow \exists U\subset M$ open with
$$\widetilde{U}=\{(p,o_p)\mid p\in U, o_p\in\{\text{orientations on }T_pM\}$$
Now I'm trying to figure out some chart $(\widetilde{U},\widetilde{\phi})$ at $(p,o_p)$ based on $(\phi, U)$ at $p$. I've tried this:
\begin{align*}
\widetilde{\phi}:\widetilde{U}&\to\mathbb{R}^n\\
(p, o_p)&\mapsto \phi(p)
\end{align*}
But that obviously doesn't work because it is not even injective. Somehow I have to involve the orientation $o_p$ in the definition, but I really don't know how to do it.
About the orientability, I guess it will have something to do with the orientability of the atlas $\{(\widetilde{U}_{\alpha}, \widetilde{\phi}_{\alpha})\}$, but since I can't figure out the definition of $\widetilde{\phi}$, I'm stuck.
Now for the connectedness of $\widetilde{M}$ and non-orientability of $M$, that I have no idea.
 A: You may want to take a look at John Lee's Introduction to Smooth Manifolds. Chapter 15 contains a fairly comprehensive discussion of orientations. In particular, the section "Orientations and Covering Maps" gives detailed answers to your three questions.
A: Almost $2$ years later, I'll give a complete answer to my own question.
Step 1 (Topology of $\widetilde{M}$): Take an atlas $\{(U_\alpha,\varphi_\alpha)\}_{\alpha\in\Lambda}$ such that $\{U_\alpha\}_{\alpha\in\Lambda}$ is a countable basis for $M$. Define the following subsets of $\widetilde{M}$:
$$U_\alpha^+:=\left\{(p,o_p)\in\widetilde{M}\mid p\in U_\alpha,\, o_p=\left[\left.\frac{\partial }{\partial\varphi_\alpha^1}\right|_p,...,\left.\frac{\partial }{\partial\varphi_\alpha^n}\right|_p\right]\right\},$$
$$U_\alpha^-:=\left\{(p,o_p)\in\widetilde{M}\mid p\in U_\alpha,\,o_p=-\left[\left.\frac{\partial }{\partial\varphi_\alpha^1}\right|_p,...,\left.\frac{\partial }{\partial\varphi_\alpha^n}\right|_p\right]\right\}.$$
We define the topology of $\widetilde{M}$ as the one generated by the basis $\{U_\alpha^+,U_\alpha^-\}_{\alpha\in\Lambda}$. This is a countable basis since $\Lambda$ is countable. In order to check that this topology is Hausdorff we only need to use the fact that the topology of $M$ is Hausdorff.
We prove, in addition, that this makes $\pi$ a continuous, open map (in fact, a double covering). Indeed, notice that for any $\alpha\in\Lambda$ we have $\pi^{-1}(U_\alpha)=U_\alpha^+\cup U_\alpha^-$ and $\pi(U_\alpha^\pm)=U_\alpha$. Since $\{U_\alpha\}_{\alpha\in\Lambda}$ is a basis for $M$ and $\{U_\alpha^+,U_\alpha^-\}_{\alpha\in\Lambda}$ is a basis for $\widetilde{M}$, consequently $\pi$ is continuous and open. Moreover, for an arbitrary $p\in M$, any open set $U_\alpha$ containing $p$ is such that $\pi^{-1}(U_\alpha)=U_\alpha^+\cup U_\alpha^-$ (disjoint union) and $\pi|_{U_\alpha^\pm}:U_\alpha^\pm\to U_\alpha$ is a homeomorphism, which shows that $\pi$ is a double covering.
Step 2 (Differentiable Structure of $\widetilde{M}$): Define $\varphi_\alpha^+:U^+_\alpha\to \varphi_\alpha(U_\alpha)\subset\mathbb{R}^n$ by $\varphi^+_\alpha=\varphi_\alpha\circ\pi|_{U_\alpha^+}$ and, similarly, $\varphi_\alpha^-:U^-_\alpha\to\varphi_\alpha(U_\alpha)\subset\mathbb{R}^n$ by $\varphi^-_\alpha=\varphi_\alpha\circ\pi|_{U_\alpha^-}$. Both $\varphi_\alpha^+,\varphi_\alpha^-$ are homeomorphisms, because $\varphi_\alpha$ and $\pi|_{U_\alpha^\pm}$ are homeomorphisms. Moreover:
\begin{align*}
\varphi_\alpha^\pm\circ(\varphi_\beta^\pm)^{-1}(x_1,...,x_n)&=\varphi_\alpha^\pm\left(\underbrace{\varphi_\beta^{-1}(x_1,...,x_n)}_{=:p},\pm\left[\left.\frac{\partial }{\partial\varphi_\beta^1}\right|_p,...,\left.\frac{\partial }{\partial\varphi_\beta^n}\right|_p\right]\right)\\
&=\underbrace{\varphi_\alpha\circ\varphi_\beta^{-1}}_{\text{smooth}}(x_1,...,x_n).
\end{align*}
(the upper indexes $\pm$ are not relevant to this argument)
This shows that the atlas $\{(U_\alpha^+,\varphi_\alpha^+),(U_\alpha^-,\varphi_\alpha^-)\}_{\alpha\in\Lambda}$ is compatible, which makes $\widetilde{M}$ a smooth manifold.
This also makes $\pi$ a local diffeomorphism, since $\pi|_{U_\alpha^\pm}=\varphi_\alpha^{-1}\circ\varphi_\alpha^\pm$ and $\varphi_\alpha,\varphi_\alpha^\pm$ are diffeomorphisms.
Step 3 (Orientability of $\widetilde{M}$): Let's construct a pointwise orientation $O:(p,o_p)\mapsto O_{(p,o_p)}$ on $\widetilde{M}$. Take an arbitrary $(p,o_p)\in\widetilde{M}$. Since $\pi$ is a local diffeomorphism, $(d\pi)_{(p,o_p)}$ is a bijective linear transformation and we may find a unique $O_{(p,o_p)}$ which corresponds to $o_p$ via $d\pi$. More precisely, define $O_{(p,o_p)}:=[(d\pi)_{(p,o_p)}^{-1}e_1,...,(d\pi)_{(p,o_p)}^{-1}e_n]$, where $\{e_1,...,e_n\}$ is any basis for $T_pM$ with $o_p=[e_1,...,e_n]$.
We show that $O$ is continuous. Notice that for a neighbourhood $U_\alpha$ of $p$, we either have $(p,o_p)\in U_\alpha^+$, in which case $O_{(q,o_q)}=\left[\left.\frac{\partial }{\partial(\varphi_\alpha^+)^1}\right|_{(q,o_q)},...,\left.\frac{\partial }{\partial(\varphi_\alpha^+)^n}\right|_{(q,o_q)}\right]$ for all $(q,o_q)\in U_\alpha^+$, or $(p,o_p)\in U_\alpha^-$, in which case $O_{(q,o_q)}=\left[\left.\frac{\partial }{\partial(\varphi_\alpha^-)^1}\right|_{(q,o_q)},...,\left.\frac{\partial }{\partial(\varphi_\alpha^-)^n}\right|_{(q,o_q)}\right]$ for all $(q,o_q)\in U_\alpha^-$. Since $(p,o_p)$ is arbitrary, this means that $O$ is continuous. Thus $\widetilde{M}$ is orientable.
Step 4 (Orientability of $M$ vs. Connectedness of $\widetilde{M}$): Suppose $\widetilde{M}$ is disconnected. Since $\pi$ is a double cover, this means that $\widetilde{M}=U\cup V$, where $U,V$ are disjoint open subsets such that both $\pi|_U:U\to M$ and $\pi|_V:V\to M$ are diffeomorphisms. As $\widetilde{M}$ is orientable, in particular $U$ is orientable, so $M$ inherits an orientation from $U$ via $\pi|_U$.
Conversely, suppose $M$ is orientable and take an oriented atlas $\{U_\alpha,\varphi_\alpha\}_{\alpha\in\Lambda}$. We show that $\widetilde{M}$ is the disjoint union of the open sets $\bigcup_\alpha U_\alpha^+$ and $\bigcup_\alpha U_\alpha^-$, which means that $\widetilde{M}$ is disconnected. Assume by contradiction that $U_\alpha^+\cap U_\beta^-\neq \emptyset$ for some $\alpha,\beta\in\Lambda$. If $(p,o_p)\in U_\alpha^+\cap U_\beta^-$, this means that $p\in U_\alpha\cap U_\beta$ with $o_p=\left[\left.\frac{\partial}{\partial \varphi_\alpha^1}\right|_p,...,\left.\frac{\partial}{\partial \varphi_\alpha^n}\right|_p\right]=$ $-\left[\left.\frac{\partial}{\partial \varphi_\beta^1}\right|_p,...,\left.\frac{\partial}{\partial \varphi_\beta^n}\right|_p\right]$, therefore $\det(D(\varphi_\alpha\circ\varphi_\beta^{-1})(\varphi_\beta(p)))<0$ (absurd, since the atlas is oriented). $_\blacksquare$
