Let $M$ be a non-orientable, connected, smooth manifold with $\text{dim }M=n$. I'm trying to fill in the ideas of a construction of a unique $2:1$ covering $\tilde{M}$ of $M$. I've pieced together parts but missing details:
Since $M$ is non-orientable, $E:=\Lambda^nM$ is a real line bundle over $M$. The smooth map $F\colon E\to\mathbb{R}$ defined by $F(v)=\|v\|^2$ has $1$ as a regular value, so $\tilde{M}:=F^{-1}(1)$ is an embedded submanifold of $E$.
Let $\pi\colon E\to M$ be the projection and $\text{pr}\colon\tilde{M}\to M$ be the restriction. This is a local diffeomorphism onto $M$ with $|\text{pr}^{-1}(p)|=2$ for any $p$.
I want to show now that $\Lambda^n\tilde{M}\cong\text{pr}^\ast E$ ($=\text{pr}^\ast\Lambda^nM$) and that $\text{pr}^\ast E$ over $\tilde{M}$ has a smooth, nowhere vanishing section. Hence $\Lambda^n\tilde{M}\to\tilde{M}$ is trivial, so $\tilde{M}$ is orientable. Moreover $\tilde{M}$ is connected giving the $2:1$ covering map $\text{pr}\colon\tilde{M}\to M$. It is also unique.
Here are my questions about the above:
(1) Why is $1$ a regular value of $F$?
(2) Why is the restriction $\text{pr}\colon\tilde{M}\to M$ a local diffeomorphism?
(3) Why is there a bundle isomorphism between $\Lambda^n\tilde{M}$ and $\text{pr}^\ast\Lambda^nM$?
(4) How to construct a smooth, nowhere vanishing section of $\text{pr}^\ast E$ over $M$?