It's not possible to put an atlas (of manifold with boundary) on $[0,1]$ because any chart whose domain contains $0$ has to have a positive derivative while any chart whose domain contains $1$ has to have a negative derivative (taking charts of $n$-manifold with boundary to have values in $\mathbb{R}_+\times\mathbb{R}^{n-1}$). Yet it has to be orientable: what's the catch?
There's an answer here: Is $[0,1]$ an *oriented* manifold with boundary? (and Stokes theorem), but I find it very distasteful. Is there a problem with defining an orientation of a manifold with boundary to be an orientation of the manifold of its interior points? Wouldn't that be a more kosher approach?
This is, afaik, the usual approach to defining an orientation on a manifold with boundary: take an orientation on the interior, which then induces an orientation on the boundary. The problem that you ran into when trying to find an oriented atlas for a manifold with boundary on $[0, 1]$ is reflective of how the induced orientation is determined. The points $0$ and $1$ end up having opposite orientations.
An orientation on an orientable manifold can be described by an ordered basis of the tangent space to a point. Say, if $(v_1, ..., v_n)$ is a basis for $T_pM,$ where $M$ is orientable, one orientation of $M$ is given by $[v_1, ..., v_n];$ the other is given by $[-v_1, ..., v_n] = [v_2, v_1, v_3, ..., v_n].$ You may have a definition of an orientation on an $n$-dimensional manifold $M$ as a nowhere-vanishing $n$-form $\omega.$ These definitions are equivalent - the orientation determined by $\omega$ is the orientation $[v_1, ..., v_n]$ iff $\omega(v_1, ..., v_n)>0.$
Then, to define the induced orientation on the boundary of $M,$ let $p\in\partial M$ and let $\varphi: U\to \mathbb{H}^n$ be a chart from a neighborhood of $p$ to the $n$-dimensional upper half-space $\mathbb{H}^n,$ i.e. $\mathbb{R}^{n-1} \times \mathbb{R}_{\ge 0}.$
Then, the orientation on $\partial M$ is $[u_1, ..., u_{n-1}],$ where $(u_1, ..., u_{n-1})$ is a basis of $T_p\partial M$ such that $[\varphi^*(0, 0, ..., 0, -1), u_1, ..., u_{n-1}]$ is the orientation on $M.$ ($\varphi^*(0, 0, ..., 0, -1)$ is the pullback of the vector $(0, ..., 0, -1),$ which can be visualized as a vector in the tangent space $T_pM$ pointing away from $T_p\partial M.$ The "outward normal vector" can also be used in place of this vector in this construction.) It takes a little bit of work to show that this is well-defined, but it is.
In your specific example, the vector $\varphi^*(0, 0, ..., 0, -1)$ points in different directions when $\varphi$ is a chart in a neighborhood of $0$ or $1,$ so the induced orientations at $0$ and $1$ are different.