I'm trying exercise 1—7(d) from Lee's Introduction to Smooth Manifolds, but unlike is claimed here: https://math.stackexchange.com/questions/2876921/maximal-smooth-atlas-for-stereographic-projection , exercise (d) actually asks something else: to show that the smooth atlases $\mathcal{A}_1 := \{(U_i^\pm, \phi_i^\pm) | i \in \{1, \ldots, n+1\}\} \text{ and } \mathcal{A}_2 := \{(S^n \setminus \{N\}, \sigma), (S^n \setminus \{N\}, \tilde\sigma)\}$ actually determine equal atlases.
Using the definitions (and/or Proposition 1.17), and using that $\mathcal{A}_1$ and $\mathcal{A}_2$ are already smooth atlases (by Example 1.31 and a proof of myself, respectively), this boils down to showing that for each $i$, the maps $$\phi_i^\pm \circ \sigma^{-1} : \sigma[S^n \setminus\{N\} \cap U_i^\pm] \to \phi_i^\pm[S^n \setminus\{N\} \cap U_i^\pm] $$ and $$\phi_i^\pm \circ \tilde\sigma^{-1} : \tilde\sigma[S^n \setminus\{S\} \cap U_i^\pm] \to \phi_i^\pm[S^n \setminus\{S\} \cap U_i^\pm] $$ are diffeomorphisms.
Indeed, then $\mathcal{A}_1 \cup \mathcal{A}_2$ is a smooth atlas, so by Prop. 1.17, they define the same smooth structure.
The involved maps took like this: $$ \begin{eqnarray} \sigma : &S^n \setminus \{N\} &\to \mathbb{R}^n\\ &(x_1, \ldots, x_{n+1}) &\mapsto \frac{(x_1, \ldots, x_{n})}{1-x_{n+1}} \end{eqnarray} $$
with inverse
$$ \begin{eqnarray} \sigma^{-1} : &\mathbb{R}^n &\to S^n \setminus \{N\}\\ &(x_1, \ldots, x_{n}) &\mapsto \frac{(2x_1, \ldots, 2x_{n}, \lVert x \rVert -1 )}{\lVert x \rVert +1} \end{eqnarray}, $$ and
$$ \begin{eqnarray} \tilde \sigma : &S^n \setminus \{S\} &\to \mathbb{R}^n\\ &(x_1, \ldots, x_{n+1}) &\mapsto \frac{(x_1, \ldots, x_{n})}{1+x_{n+1}} \end{eqnarray} $$ and transition function $$ \begin{eqnarray} \tilde \sigma \circ \sigma^{-1} : &\mathbb{R}^n &\to \mathbb{R}^n\\ &(x_1, \ldots, x_{n}) &\mapsto \frac{(x_1, \ldots, x_{n})}{\lVert x \rVert} \end{eqnarray}, $$ and, for $x^i = \sqrt{1 - x_1^2 - \ldots \hat{x_i^2} - \ldots x_{n+1}^2} ,$
$$ \begin{eqnarray} \phi_i^\pm : &U_i^\pm \cap S^n \setminus \{N\} &\to \mathbb{B}^n\\ &(x_1, \ldots, x_{n+1}) &\mapsto \pm(x^1, \ldots, \hat x^i, \ldots, x^{n+1}) \end{eqnarray} $$ On the one hand, I could write, for $\sigma(u)$ an arbitrary element of the domain, $$ \phi_i^\pm \circ \sigma^{-1} (\sigma(u)) = \phi_i^\pm(u) = (u^1, \ldots, \hat{u^i}, \ldots, u^{n+1}),$$
but that kind of trivialises the matter.., while on the other hand we could write out explicitly
$$ \phi_i^\pm \circ \sigma^{-1} (x) = \phi_i^\pm\left(\frac{(2x_1, \ldots, 2x_{n}, \lVert x \rVert - 1)}{\lVert x \rVert + 1}\right) = \pm\left(\sqrt{1 - \frac{4x_2^2 + 4x_3^2 +\ldots + 4x_{n}² + (\lVert x \rVert -1)^2} {(\lVert x \rVert +1)^2} } , \ldots, \hat{x^i}, \ldots, x^{n+1}\right),$$ which gets extremely cumbersome.
Am I getting at the correct approach here? Could anyone help me along? I am not sure whether I am getting the key ideas and the right interpretations of the notations here, and am kind of lost.
Your approach is correct, but you made some mistakes.
Let us agree to write points of $\mathbb R^n$ in the form $x = (x_1,\ldots,x_n)$ and points of $S^n$ in the form of pairs $(x,x_{n+1})$ with $x \in \mathbb R^n$.
Then the stereographic projections $\sigma : S^n \setminus \{N\} \to \mathbb R^n$ and $\tilde \sigma : S^n \setminus \{S\} \to \mathbb R^n$ are given by $$\sigma(x,x_{n+1}) = \frac{1}{1-x_{n+1}} x,\\ \tilde \sigma (x,x_{n+1}) = \frac{1}{1+x_{n+1}} x$$ and their inverses by $$\sigma^{-1}(x) = \frac{1}{\lVert x \rVert^2 +1} (2x, \lVert x \rVert^2 -1 ) ,\\ \tilde \sigma^{-1}(x) = \frac{1}{\lVert x \rVert^2 +1} (2x, 1 -\lVert x \rVert^2 ) . $$ Domain and range of the transition functions $\tilde \sigma \circ \sigma^{-1}$ and $\sigma \circ \tilde \sigma^{-1}$ are not $\mathbb R^n$, but $\mathbb R^n \setminus \{0\}$. In fact, the intersection of the domains of the charts $\sigma$ and $\tilde \sigma$ is $$D = (S^n \setminus \{N\}) \cap (S^n \setminus \{S\}) = S^n \setminus \{N, S\} .$$ Since $\sigma(S) = \tilde \sigma(N) = 0$, the domain of the transition function $\tilde \sigma \circ \sigma^{-1}$ is $\sigma(D) = \mathbb R^n \setminus \{0\}$ and its range is $\tilde \sigma(D) = \mathbb R^n \setminus \{0\}$. The same is true for $\sigma \circ \tilde \sigma^{-1}$. Both transition functions functions are given by $$x \mapsto \frac{x}{\lVert x \rVert^2} . $$ You wrote that you checked the smoothness of the transition functions (which is true), but perhaps you used the wrong functions.
Note that the formulae for the inverse stereographic projections actually give us smooth functions $$\rho : \mathbb R^n \to \mathbb R^{n+1}, \rho(x) = \frac{1}{\lVert x \rVert^2 +1} (2x, \lVert x \rVert^2 -1 ) ,\\ \tilde \rho : \mathbb R^n \to \mathbb R^{n+1}, \tilde \rho(x) = \frac{1}{\lVert x \rVert^2 +1} (2x, 1 -\lVert x \rVert^2 ) . $$
Similarly the formulae for the stereographic projections give us smooth functions $$\Sigma : \mathbb R^{n+1}_+ \to \mathbb R^n, \Sigma(x) = \frac{1}{1-x_{n+1}} x ,\\ \tilde \Sigma : \mathbb R^{n+1}_- \to \mathbb R^n, \tilde \Sigma(x) = \frac{1}{1+x_{n+1}} x $$ defined on the open subsets $\mathbb R^{n+1}_\pm = \{ (x_1,\ldots,x_{n+1}) \in \mathbb R^{n+1} \mid x_{n+1} \ne \pm 1\}$.
The $\phi_i^\pm : U_i^\pm \to \mathbb B^n$ are given by $$\phi_i^\pm(x_1,\ldots,x_{n+1}) = (x_1,\ldots,\hat x_i,\ldots, x_{n+1}) .$$ No sign occurs here. Note that $\phi_i^\pm$ is the restriction of the projection $p_i : \mathbb R^{n+1} \to \mathbb R^n, p_i(x_1,\ldots,x_{n+1}) = (x_1,\ldots,\hat x_i,\ldots, x_{n+1})$ which is clearly smooth.
Their inverses are given by $$(\phi_i^\pm)^{-1} : \mathbb B^n \to U_i^\pm,(\phi_i^\pm)^{-1}(x_1,\ldots,x_n) = (x_1,\ldots,x_{i-1}, \pm\sqrt{1 - \lVert x \rVert^2},x_i,\ldots,x_n) .$$ We can again regard them as functions $\psi_i^\pm : \mathbb B^n \to \mathbb R^{n+1}$ which are clearly smooth.
Let us determine the transition functions between $\sigma$ and the $\phi_i^\pm$ (for the chart $\tilde \sigma$ it works similarly).
We have $$(S^n \setminus \{N\}) \cap U_{n+1}^+ = U_{n+1}^+ \setminus \{N\}$$ and $$(S^n \setminus \{N\}) \cap U_i^\pm = U_i^\pm$$ in all other cases.
The function $\phi_i^\pm \circ \sigma^{-1} : \sigma ((S^n \setminus \{N\}) \cap U_i^\pm) \to \phi_i^\pm((S^n \setminus \{N\}) \cap U_i^\pm)$ is given by $$(\phi_i^\pm \circ \sigma^{-1})(x) = (p_i \circ \rho)(x) .$$ By the ordinary chain rule of multivariable calculus $p_i \circ \rho$ is smooth, thus $\phi_i^\pm \circ \sigma^{-1}$ is smooth because it is the restriction of a smooth map to an open subset of $\mathbb R^n$.
The function $\sigma \circ (\phi_i^\pm)^{-1} : \phi_i^\pm((S^n \setminus \{N\}) \cap U_i^\pm) \to \sigma ((S^n \setminus \{N\}) \cap U_i^\pm)$ is given by $$(\sigma \circ (\phi_i^\pm)^{-1} )(x) = (\Sigma \circ \psi_i^\pm)(x) .$$ As above we see that this function is smooth.