From Hatcher:
In the normal bundle $N S^n$ the identification $(x, v) ∼ (−x, −v)$ can be written as $(x,tx) ∼ (−x,t(−x))$. This identification yields the product bundle $\mathbb{RP}^n×\mathbb{R}$ since the section $x \to (−x, −x)$ is well-defined in the quotient.
What does this even mean? Is the section $S^n \to NS^n$ or $\mathbb{RP}^n \to N \mathbb{RP}^n$? What does the second one mean? I understand that the normal bundle on $S^1$ is trivial, and I see how when $n = 1$ this is different from the Mobius bundle, but I don't understand what Hatcher is trying to say here. Any insight would be appreciated.
If $M$ is a submanifold of $\mathbb{R}^n$, the normal bundle of $M$ is the vector bundle $NM$ such that for every $x\in M$, $NM_x$ the fibre of $NM$ at $x$ is the subspace of $\mathbb{R}^n$ othogonal to $T_xM$.
If $M=S^n$, $NS_x=(x,tx),t\in\mathbb{R}$. You cannot define the normal bundle of $\mathbb{R}P^n$ since it is not embedded in $\mathbb{R}^n$, but $NS^n$ behaves well in respect to $(x,tx)\rightarrow (-x,-tx)$ as Hatcher mentioned and induces a bundle on $\mathbb{R}P^n$ which is trivial. Basically, you can lift the action of $\mathbb{Z}/2$ of on $S^n$ and the quotient $NP^n$ is a trivial bundle. Another way to see this, is t consider the map $q:\mathbb{R}P^n\times \mathbb{R}\rightarrow NP^n$ defined by $q([x],t)\rightarrow p(x,tx)$ where $p:S^n\rightarrow \mathbb{R}P^n$ is the canonical projection and $p(x)=[x]$. In fact Hatcher says that $p(x,t)=p(-x,t)$ thus $q$ is well defined.