Ok this is already quite a mouth full, so let me try to give answers to some of your questions:
The main issue is that Sobolev mappings are defined via a boundedness concept (you ask for $L^p$-integrability conditions) and boundedness is not an intrinsic concept on a manifold. This means that definitions in charts usually blow up when boundedness is concerned (as I can always compose with an arbitrary diffeomorphism to make new charts and by choosing badly I can blow up every such condition). This is often glossed over in the older Sobolev literature on manifolds (or they define things chart independent in which case the problem does not exist). For most of what follows now, I refer to the great memoir by Inci, Kappeler and Topalov: On the regularity of the composition of diffeomorphisms (arxiv version is here [1]) where these things are worked out.

1. Atlases with the desired property always exist. Indeed one can even strengthen the conditions to make the atlases more convenient to work with (e.g. Lipschitz boundary so restrictions of your Sobolev maps behave nicely). Atlases with the additional properties are called fine cover in [1] (see Section 3.1 in [1] for discussion and an existence proof)  
 
2. These atlases are necessary to talk about Sobolev functions on a manifold **in charts**. There are chart independent definitions which usually require you to pick a Riemannian metric as an additional structure to make sense of integration and boundedness. For example the classical book by Palais: Foundations of Global Non-Linear Analysis, [2] does it this way. See the Appendix of [3] for a comparison of the notion (even in the boundary case).
3. For the counterexample we can look at [1] p.43 (which I am quoting here):
Consider the torus $M= \mathbb{R}/\mathbb{Z}$ and let $f\colon (−1/2, 1/2) \rightarrow \mathbb{R}$ be the function
$$f(x) := \begin{cases} x^{2/3} &, x \in [0, 1/2)\\ 
 (−x)^{2/3} &, x \in [−1/2, 0).
\end{cases}$$
Extending $f$ periodically to $\mathbb{R}$ we get a function on $M$ that we denote by the same letter. It is not hard to see that $f \in H^1(M, \mathbb{R})$. Now, introduce a new coordinate $y = x^2$ on the open set $(0,1/2) \subseteq M$. Then $\tilde{f}(y) := f(x(y)) = y^{1/3},\ y \in (0, 1/4)$. We have, $\tilde{f}'(y) = 1/(3y^{2/3})$, and hence, $\tilde{f}′ \not \in L^2((0, 1/4), R)$. This shows that $\tilde{f} \not \in H^1((0, 1/4), R)$.
This is an easy counterexample showing that a function may fail to be Sobolev in certain coordinate charts (and the boundedness conditions remedy that). Similar ideas will also produce differential forms which are $H^s$ in one but not the other coordinate system.

This answers half of your questions (since I am no expert on Sobolev spaces I will leave the other questions to somebody who is more in the know).


[1]: https://arxiv.org/abs/1107.0488
[2]: http://vmm.math.uci.edu/PalaisPapers/FoundationsOfGlobalNonlinearAnalysis.pdf
[3]: https://arxiv.org/abs/1909.09982