I am writing about the following theorem:
7.2.1. Theorem. If $\mathcal R$ is a von Neumann algebra acting on a Hilbert space $H$ and $x_0$ is a unit generating vector for $M$, then, for each vector $z_0$ in $H$, there are operators $T$ and $S$ in $\mathcal R$ and a vector $y_0$ orthogonal to the null space of $T$ such that $Ty_0=x_0$ and $Sy_0=z_0$.
The first paragraph of in this proof is completely clear.
In the second paragraph, a sequence a operators is introduced which makes the assertion is proved. I strongly would like to know how they have found if such a sequence is considered, then the assertion will be obtained!
In the first paragraph of the proof, the sequence $\{T_n\}$ is constructed with $\sum T_nx_0=z_0$. There is no reason for the series $\sum T_n$ to converge; so the idea is to get $T$ and $y_0$ such that $Ty_0=x_0$ and $\sum T_nT$ converges.
The point $y_0$ is chosen as a wot cluster point of a sequence $\{H_nx_0\}$. If one gets $\{H_n^{-1}\}$ to be convergent to a $T$, then such $T$ will satisfy $Ty_0=x_0$.
We also need $\sum T_nT$ to be convergent. One way to guarantee this is to have $\|T_nT\|\leq2^{-n}$, which we can write as $$\tag{1} 4^nT^*T_n^*T_nT\leq I. $$ Since we want $T=\lim H_n^{-1}$, for $(1)$ to hold we need $$ 4^kH_n^{-1}T_k^*T_kH_n^{-1}\leq I $$ when $n\geq k$, or $$\tag{2} 4^kT_k^*T_k\leq H_n^2. $$ This suggests taking $H_n^2=\sum_{k=0}^n4^kT_k^*T_k$. Since we want them to be invertible, we define $$ H_n^2=I+\sum_{k=0}^n4^nT_k^*T_k. $$ It is not clear a priori what happens with the norms of these $H_n$, but since $H_n\geq I$ and $I\leq H_n\leq H_{n+1}$, the sequence $\{H_n^{-1}\}$ is positive and decreasing, and so it has a sot limit $T$.
From $H_nx_0\to y_0$ (with a subsequence), one obtains $Ty_0=x_0$.