When you say you have no problems visualising the stereographic projections you (probably) see the sphere as embedded in $\Bbb R^3$.
To define a chart you have to tell which point in $U$ goes where. You can't do this unless you give the points names a priori to having access to a chart. In the special case where the manifold in question lives inside some $\Bbb R^N$ ($N > n$) it might be convenient to use their Cartesian coordinates in $\Bbb R^N$ as such names. Once you defined a chart you can use its inverse to refer to the manifold points. Now you are left with an "$n$-dimensional namespace": an $n$-tuple is enough to reference a point. If a point $p \in \Bbb R^n$ references a manifold point $q$ via a chart, then there is a neighbourhood of $p$ in $\Bbb R^n$ which is diffeomorhic to a neighbourhood of $q$ in $M$. By contrast, in each neighbourhood of $q$ in $\Bbb R^N$, there are point, which are not used as names for manifold points. That's why $N$ doesn't say anything about the dimension of your manifold, whilst $n$ is actually used to define the dimension of $M$.
"When people write something like local coordinates $x_1,…,x_n$ for the manifold M", they usually refer to a single chart $x: U \to U' \subset \Bbb R^n$ and use $x_1, \dots x_n$ as abbreviatons to the projections of $x$ to it's component in $\Bbb R^n$, precisely: $x_i := pr_i \circ x$, where $pr_i: \Bbb R^n \to \Bbb R$ is the projection to the $i$th component relative to the standard coordinates of $\Bbb R^n$. So it is actually the chart combined with what you do, when you use coordinates in linear algebra.
Unfortunatly I can't help you with the part of your question concerning the contact structure. But as I understand the link you gave, you missed some termes in the definition of the standard contact strucure? Because $S^3$ lives in $\Bbb R^4$, so you can name it's points after their cartesean coordinates $(x_1, y_1, x_2, y_2)$ and then the contact structure would in these cartesean coordinates be given as $\alpha_0 = x_1 d y_1 - y_1 d x_1 + x_2 d y_2 - y_2 d x_2$.
A mapping $U \to \mathbb R^n$ from an open subset $U$ of your manifold
is not a chart if one point of $U$ is "wrapped around the edges."
You may have answered your own question with that single sentence.
The problem isn't that it's impossible to make a map projection covering every point on Earth--the problem is that every such projection represents at least one point too many times. It gives a one-to-many correspondence that is not a function.
Alternatively, if you erase the redundant copies of each such point and keep just one, you still don't have a continuous mapping at that point, so you still don't have a chart in the mathematical sense.
If you remove one point from a sphere $S$ to get an open set $S_1,$
a stereographic projection gives a nice conformal mapping $S_1 \to \mathbb R^2.$
In mathematics, however, a sphere with one point removed is not a sphere.
It's a sphere with one point removed.
Best Answer
Let $(x, y)$ denote Cartesian coordinates on the plane. Smooth homeomorphisms such as \begin{align*} \phi_{0}(x, y) &= (x, y), \\ \phi_{1}(x, y) &= (x^{3}, y), \\ \phi_{2}(x, y) &= (x, y^{3}), \\ \phi_{3}(x, y) &= (x^{3}, y^{3}), \end{align*} among an infinite-dimensional space's worth of other examples, define pairwise-incompatible charts on the plane.
To get charts on the sphere, it suffices to pick two of these and follow them by a compatible pair of charts for the sphere, such as stereographic projection from the north pole and stereographic projection from the south pole.
The conceptual points are:
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One doesn't normally see this type of example mentioned explicitly: It's typical behavior for charts of topological manifolds (homeomorphisms are "rarely" diffeomorphisms), and inapplicable to the study of smooth manifolds (no smooth atlas contains two incompatible charts).