According to the Loring W. Tu's book "An Introduction to Manifolds", we have a standard oriented atlas on $[a,b]$, with the charts $([a,b[, \phi)$ and $(]a,b],\psi)$, where $\phi \colon [a,b[ \longrightarrow [0,b-a[$ is given by $\phi (t) = t-a$ and $\psi \colon ]a,b]\longrightarrow ]a-b,0]$ given by $\psi (t) = t-b$. Also, $\psi$ is considered an atlas, since for 1-dimensional manifolds with boundary, we allow both $\mathbb{H}^1 := \mathbb{R}_{\geq0}$ and $\mathbb{L}^1 := \mathbb{R}_{\leq0}$ as models, which allows us to call a closed interval in $\mathbb{R}$ an oriented 1-dimensional manifold with boundary. I show below a paragraph in the book in which the author discusses the outward-pointing vectors on $[a,b]$.
One of the things I'm possibly not understanding fully is why he says we have a standard coordinate $x$ in the interval (since we need two charts to cover it) and consequently, why he says that we have a vector field $\frac{d}{dx}$ on the manifold and an orientation form $dx$. But I think of that, maybe wrongfully so, as minor details which are left for the reader to fill.
The "main issue" I'm having with this example is the fact that, in this example, as the author states, the outward-pointing vector on $b$ is $\frac{d}{dx}$, which has a positive component associated with $\frac{d}{dx}$, seemingly contradicting a previous statement which he had made (shouldn’t it have a negative component?). I will leave a screenshot of the statement below. I tried to work out the details in this example, but I ran into the same problem.
Where am I think wrong? I appreciate all the help in advance.
$x$ is really just the identity map on $[a,b]$. So $x\circ \phi^{-1} = \phi^{-1}, x \circ \psi^{-1} = \psi^{-1}, \phi \circ x^{-1} = \phi, \psi \circ x^{-1} = \psi$. All of these maps are smooth, which means that $x$ is an embedding of $[a,b]$ into $\Bbb R$. So, yes, it is a coordinate on $[a, b]$, and quite obviously, a natural one. The description given of $\frac d{dx}$ is accurate. At $b, \frac d{dx}$ does point outward, and at $a$, it is $-\frac d{dx}$ that points outward. What previous statement do you think this contradicts?
The only reason you need two charts here is that for technical reasons, the charts on manifolds with boundaries are into half-spaces, so a single chart can only handle one of the ends. For $1$-dimensional manifolds, there would be no problem with allowing a chart to be into a compact interval as well. But there is no reason to make this special exception to the normal definition, as it doesn't add anything that wasn't already available.
I think I see your problem now: You are concerned about this:
where $p$ is on the boundary, and in local coordinates at $p, X_p = \sum_i a^i(p)\left.\frac{\partial}{\partial x^i}\right|_p$.
In this, they are specifically talking about a chart carrying $p$ to the boundary of $\Bbb H^n$, not $\Bbb L^n$. $a^n(p) < 0$ for outward pointing vectors because for points on the boundary of $\Bbb H^n$, it is $-\frac{\partial}{\partial x^n}$ that points out of $\Bbb H^n$. For $\Bbb L^n$, the opposite is true. $\frac{\partial}{\partial x^n}$ points outward and $-\frac{\partial}{\partial x^n}$ points inward.
By section 22.5, they are no longer talking about oriented manifolds, and have slipped back into describing unoriented manifolds, where it is sufficient to have charts into subsets of $\Bbb H^n$ exclusively. As such, they did not think about the need to have charts into $\Bbb L^n$ to have a compatible orientation.