Understanding the orientable double cover

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 $\widetilde{M}$: $$U_\alpha^+:=\left\{(p,o_p)\in\widetilde{M}\mid o_p=\left[\left.\frac{\partial }{\partial\varphi_\alpha^1}\right|_p,...,\left.\frac{\partial }{\partial\varphi_\alpha^n}\right|_p\right]\right\}$$ $$U_\alpha^-:=\left\{(p,o_p)\in\widetilde{M}\mid o_p=-\left[\left.\frac{\partial }{\partial\varphi_\alpha^1}\right|_p,...,\left.\frac{\partial }{\partial\varphi_\alpha^n}\right|_p\right]\right\}$$

The topology of $\widetilde{M}$ will be the one generated by the basis $\{U_\alpha^+,U_\alpha^-\}_{\alpha\in\Lambda}$. This is a countable basis since $\Lambda$ is countable and it's easy to check this topology is Hausdorff by the fact that $M$ is Hausdorff.

Furthermore, this makes $\pi$ continuous and open (in fact, a double covering): 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. Besides, for an arbitrary $p\in M$, a neighbourhood $U_\alpha$ 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 means $\pi$ 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)

So the atlas $\{(U_\alpha^+,\varphi_\alpha^+),(U_\alpha^-,\varphi_\alpha^-)\}_{\alpha\in\Lambda}$ is compatible and $\widetilde{M}$ is a smooth manifold.

Besides, this 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, so there is a unique $O_{(p,o_p)}\in T_{(p,o)}\widetilde{M}$ which is mapped to $o_p$ via $d\pi$. Explicitly, $O_{p,o_p}=[(d\pi)_{(p,o_p)}^{-1}e_1,...,(d\pi)_{(p,o_p)}^{-1}e_n]$, where $\{e_1,...,e_n\}$ is any basis for $T_pM$ with $o_p=[e_1,...,e_n]$.

Now 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 $O$ 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 $\widetilde{M}=U\cup V$, where $U,V$ are disjoint open subsets such that both $\pi|_U:U\to M$ and $\pi|_V:V\to M$ are diffeomorphisms. Because $\widetilde{M}$ 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 will show that $\widetilde{M}$ is the disjoint union of the open sets $\bigcup_\alpha U_\alpha^+$ and $\bigcup_\alpha U_\alpha^-$, which means $\widetilde{M}$ 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 $p\in U_\alpha\cap U_\beta$ 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). $_\blacksquare$

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.