As you might recall, every smooth atlas $\mathcal{A}$ for a topological manifold $M$ is contained in a $\textbf{unique maximal smooth atlas}$ ($\textit{Proposition 1.17}$ ). So by showing that $\mathcal{A} = \{(\Bbb{S}^n \smallsetminus N,\sigma), (\Bbb{S}^n\smallsetminus S,\tilde{\sigma})\}$ is a smooth atlas for $\Bbb{S}^n$, you automatically have a unique smooth structure determined by $\mathcal{A}$. Put it another way, $\mathcal{A}$ defines a smooth structure on $\Bbb{S}^n$. So it's not a typo.
The answer is positive. The maximal atlas containing an arbitrary atlas is obtained by adding all the charts that are compatible with the atlas. As pointed out by Gnampfissimo in the comment, the crux of the problem lies in showing that if two charts, say $(V_1,\psi_1 )$ and $(V_2,\psi_2 )$, are compatible with an atlas $\mathfrak{U}$, then $(V_1,\psi_1 )$ and $(V_2,\psi_2 )$ are compatible with each other. So let us try to prove this.
Let $p\in V_1\cap V_2$. We would like to show that $\psi_2\circ\psi_1^{-1}:\psi_1(V_1\cap V_2)\rightarrow\psi_2(V_1\cap V_2)$ is smooth at $\psi_1(p)$. To prove this, choose a chart $(U,\phi )$ in $\mathfrak{U}$ that contains $p$. Then both $\phi\circ\psi_1^{-1}:\psi_1(U\cap V_1)\rightarrow\phi(U\cap V_1)$ and $\psi_2\circ\phi^{-1}:\phi(U\cap V_2)\rightarrow \psi_2(U\cap V_2)$ are smooth. So there are
- neighborhoods $O_1$, $O_2$ of $\psi_1(p)$ and $\phi(p)$, respectively, in $\mathbb{R}^n$,
- a smooth map $f:O_1\rightarrow\mathbb{R}^n$ that agrees with $\phi\circ\psi_1^{-1}$ on $O_1\cap\psi_1(U\cap V_1)$, and
- a smooth map $g:O_2\rightarrow\mathbb{R}^n$ that agrees with $\psi_2\circ\phi^{-1}$ on $O_2\cap\phi(U\cap V_2)$.
By taking the intersection of $O_1$ with $f^{-1}(O_2)$ if necessary, we may assume here that $O_1$ satisfies $f(O_1)\subset O_2$. Then $g\circ f:O_1\rightarrow \mathbb{R}^n$ is a smooth map from a neighborhood of $\psi_1(p)$ to $\mathbb{R}^{n}$. So if we can show that $g\circ f$ agrees with $\psi_2\circ\psi_1^{-1}$ on $O_1\cap\psi_1(V_1\cap V_2)$, then we can say that $\psi_2\circ\psi_1^{-1}:\psi_1(V_1\cap V_2)\rightarrow\psi_2(V_1\cap V_2)$ is smooth at $\psi_1(p)$, and we are done.
For sure, $g\circ f$ agrees with $\psi_2\circ\psi_1^{-1}$ on $O_1\cap\psi_1(U\cap V_1\cap V_2)$. But what about on the set $\{O_1\cap \psi_1(V_1\cap V_2)\}- \{O_1\cap \psi_1(U\cap V_1\cap V_2)\}$? We have no information whatsoever about the behavior of $g\circ f$ on this set. What to do?
Here's a way to proceed: Since $\psi_1(U\cap V_1 \cap V_2)$ is open in $\psi_1(V_1\cap V_2)$, there is an open set $A$ in $\mathbb{R}^n$ such that $A\cap \psi_1(V_1\cap V_2)=\psi_1(U\cap V_1 \cap V_2)$.
Then we have $$(O_1\cap A)\cap \psi_1(U\cap V_1\cap V_2)=(O_1\cap A)\cap \psi_1 (V_1\cap V_2),$$
so the replacement of $O_1$ by $O_1\cap A$ fixes everything.
Best Answer
You can simply define the transition maps, once the atlas is given.
There is a transition map which I shall denote $\psi_{m,n}$ for every pair of indices $m,n$ having the property that $U_m \cap U_n \ne \emptyset$.
The domain of $\psi_{m,n}$ is the set $\phi_m(U_m \cap U_n) \subset \mathbb R^k$ (I'm assuming implicitly that $k$ is the dimension of the manifold).
The range (or codomain) of $\psi_{m,n}$ is the set $\phi_n(U_m \cap U_n) \subset \mathbb R^k$.
And the formula for $\psi_{m,n} : \phi_m(U_m \cap U_n) \to \phi_n(U_m \cap U_n)$ is $$\psi_{m,n}(p) = \phi_n(\phi^{-1}_m(p)), \quad p \in \phi_m(U_m \cap U_n) $$
Also, once all of this is written down, one can use the definition of a manifold together with the Invariance of Domain Theorem to prove that the domain and range of $\phi_{m,n}$ are both open subsets of $\mathbb R^k$, and one can show that $\psi_{n,m}$ is an inverse map of $\psi_{m,n}$, hence each transition map is a homeomorphism from its domain to its range.
And once that is done, you can now ask yourself questions that are aimed at determining whether your manifold is a $C^\infty$ manifold, or a $C^2$ manifold, or a $C^1$ manifold or whatever smoothness property you want. Namely: Are the functions $\{\psi_{m,n}\}$ all $C^\infty$? or are they all $C^2$? or $C^1$?