Equality of sum and direct sum of projection operators I am reading the book "An introduction to Operator Algebras" by Kehe Zhu and I am confused by the proof of the following lemma:

$\textbf{Lemma 26.2}$ Let $\mathcal A$ be a von Neumann Algebra. The sum of centrally orthogonal abelian projections $\{P_i\}_{i\in I}$ in $\mathcal A$ is again abelian.  

The proof starts like the following:  
For each $i\in I$ let $Z_i := Z(P_i)$ be the central support of $P_i$ [that is, the smallest central projection projection majarizing/bigger than $P_i$]. Put $P:= \sum_{i\in I} P_i$. Since $P_i \leq Z_i$ and the $Z_i$'s are mutually orthogonal, we have $Z_iP_j =0$ and $P_i = Z_i P$ for all $i \neq j$. For each $T\in \mathcal A$
$$PTP = \left( \sum_{i\in I} Z_i PT\right) P = \sum_{i\in I} Z_iPTZ_iP = \bigoplus_{i\in I}Z_i PT Z_i P = \bigoplus_{i\in I}P_i T P_i.$$
I don't understand these last calculations. Why does the "normal" sum equal the direct sum? As far as I understand this, the direct sum acts on the Hilbert space $\bigoplus_{i\in I}\mathcal H$ while the normal sum acts on $\mathcal H$ so I don't see how equality there makes sense. Apart from that I don't understand the role of the $Z_i$ here, they should be the reason for equality. Can anyone enlighten me?
Some notes on terminology:
$1)$ "Projection" always means an orthogonal projection operator in the von Neumann Algebra $\mathcal A \subset B(\mathcal H)$,
$2)$ A projection $P \in \mathcal A$ is called abelian if and only if $P\mathcal A P$ is abelian,
$3)$ The sums are always taken to be in the strong operator topology on $B(\mathcal H)$,
$4)$ The $Z_i$ are defined as the smallest central projections such that $Z_i \geq P_i$, central meaning that $Z_i$ commutes with every operator in $\mathcal A$.
 A: The questions are two:


*

*Why does the symbol $\bigoplus$ appear in the sum?

*What is the role of the $Z_i$?


To answer lets go through the proof. What we want to do is to see that $P\mathcal A P$ is an abelian algebra. The proof starts by considering the expression $PTP$ for $T\in\mathcal A$ arbitrary. Now the $P$ on the left and the $P$ on the right can be expanded to get
$$PTP = (\sum_i P_i) T(\sum_j P_j)$$
where the sums converge in SOT.
It would be ideal if we could unify both sums to one summand, if that were the case then $PTP$ would be equal to $\sum_i (P_iTP_i)$ where the $P_i$ are mutually orthogonal projections and the sum converges in SOT. This means that the operator $PTP$ may be decomposed as a direct sum of operators $PTP = \bigoplus_i P_i TP_i \oplus 0_{P^\perp}$, where the individual $P_i$ have domain and image $\mathrm{im}(P_i)$ and $0_{P^\perp}$ is the zero operator on the orthogonal complement of $\bigoplus \mathrm{im}(P_i)$ in $\mathcal H$. This decomposition may then be done for the entire algebra: $P\mathcal A P = \bigoplus_i P_i \mathcal AP_i$ so $P\mathcal AP$ is a direct sum of abelian algebras and as such also abelian.
The $Z_i$ enter the proof as the mechanism that can perform this reduction from two sums to one sum. Here the properties $Z_i P_i =P_i$, $Z_iZ_j=0$ for $i\neq j$ and $Z_i$ being central are what we use. Then one calculates:
$$\sum_{ij} P_i T P_j = \sum_{ij} Z_iP_i T Z_j P_j = \sum_{ij} \underbrace{Z_iZ_j}_{=\delta_{ij}Z_i}\, P_i T P_j = \sum_i Z_iP_i TP_i = \sum_i P_iTP_i.$$
