This is a theorem with a direct proof by construction.
The statement simply says "there exists a rearrangement". So the book assembles one. And it does not mean there cannot exist any other construction! The proof states one case, namely expression (25) and shows it works. That's it.
The structure of the proof is basically the following.
First, we put up Expression (25). Then we show that all its constituents indeed exist and make sense, namely $p_n, q_n, P_{m_k}, Q_{m_k}, \beta_n, \alpha_n$.
A picture is worth a thousand words.
Curves in blue and in green are sequences $\beta_n$ and $\alpha_n$, they converge to $\beta$ and $\alpha$ respectively.
Curve in red is our Expression (25). The way we painted this picture shows a few things about Expression (25).
The first terms of it are positive, so the reason why $\beta_1 >0$, and the curve first goes up. Again, in our construction we first put Expression (25) as a basis, starting point of the proof. The first terms of it are positive, so $\beta_1$ must be positive. We tweak betas and other components to make them tie in with Expression (25).
Next, $x_n > \beta_n$ and $y_n < \alpha_n$, so the red curve of Expression (25) sticks out a bit "outside" sequences $\beta_n$ and $\alpha_n$. By one last positive term and negative term $P_{m_n}$ and $Q_{k_n}$.
Differences between Expression (25) and these sequences (marked X) become smaller because $P_n \to 0$ and $Q_n \to 0$. It is precisely for this reason it becomes finally "clear" that Expression (25) cannot converge to any number greater than $\beta$ or smaller than $\alpha$.
Can we have an alternative construction? Why not!
Just change Expression (25), and tweak all other elements accordingly, e.g. make Expression (25) "expand" inside the "space" between $\beta$ and $\alpha$. Make the first terms negative numbers, the curve would first go down in this case. The differences (marked Y) would also gradually become smaller. And the whole thing would work as well.
You might ask why the theorem is stated only for the real numbers? That's because all inequalities do not make sense for the complex numbers, which do not have '<' defined.
I hope the following answers (part of) your question.
If $M,N$ are von Neumann algebras and if we have bounded nets $\{m_i\}_i\subseteq M$ and $\{n_i\}_i\subseteq N$ such that $m_i \to m$ $\sigma$-weakly on $M$ and $n_i\to n$ $\sigma$-weakly on $N$, then it is true that
$$m_i\otimes n_i\to m\otimes n$$
$\sigma$-weakly in $M \overline{\otimes} N$.
Proof: We may assume $M\subseteq B(H), N \subseteq B(K)$. Since any normal functional on $M \overline{\otimes}N$ is in the norm-closure of the linear span of the functionals $\omega_{\xi, \eta} \overline{\otimes} \omega_{\xi',\eta'}$, it suffices to show (via a density argument, using the fact that the nets are bounded) that
$$(\omega_{\xi, \eta} \overline{\otimes} \omega_{\xi',\eta'})(m_i\otimes n_i)\to (\omega_{\xi, \eta} \overline{\otimes} \omega_{\xi',\eta'})(m\otimes n)$$
which is evident, given the assumption. Alternatively, if you know that on bounded subsets, the weak and the $\sigma$-weak topology coincide, you can simplify the above a bit.
In your situation, we can apply this to $M= N= \mathcal{A}^{**}$
to conclude that
$$f_p^n \otimes f_q^n \to p\otimes q$$
in $\mathcal{A}^{**}\overline{\otimes}\mathcal{A}^{**}$.
Best Answer
Your attempt looks fine, except for the last part (note though, that $\|Ax\|^2=\sum_{n=1} |\alpha_n| \|P_n x\|^2$ by our own argumentation - you don't just have inequality). It's not clear what $\mathcal{S}$ is, and there is no way that $Ax=\sum_{n=1}^{\infty} \alpha_n x$ for any $x$ since the $\alpha_n$ are pairwise orthogonal.
However, if $x\in Ran P_n$, then $A x= \alpha_n x$ so if you pick $x\in Ran P_n$ with $\|x\|=1$, you get that $\|A\|\geq \alpha_n$ for every $n$, which establishes the other inequality.
Finally, to prove norm convergence when $\alpha_n\to 0$, let $A_N:=\sum_{n=1}^N \alpha_n P_n$ and note that $$ \|(A-A_N) x\|^2=\sum_{n=N+1}^{\infty} |\alpha_n|^2 \|P_n x\|^2\leq \sup_{n>N} |\alpha|_n^2 \|x\|^2, $$ which implies that $\|A-A_N\|\to 0$ by our assumption of decay.