Let me use capital letter $W$ instead of small letter $w$ (as $W$ is actually a matrix).
So, let $W$ be the matrix that contains in row $i$ and column $j$ the probability that $j$ hears from $i$. Then average number of pieces of information that $j$ gets from $i$ in one period is given by $W(i,j)$.
The average number of information that $j$ gets from $i$ that takes 2 periods to travel is the information that goes first from $i$ to some intermediate person $k$ and then from $k$ to $j$, via the path $i \to k \to j$. The amount of information transmitted through this path is $W(i,k) \cdot W(k,j)$. To see this, notice that $W(i,k)$ is the amount of info send to $k$ and of this a fraction $W(k,j)$ is sent forward to $j$. Now, this has to be summed over all intermediate nodes $k$ (i.e. all paths of length 2). so we get:
$$
\sum_{k} W(i,k) \cdot W(k,j) = (W \times W)_{i,j} = (W^2)_{i,j}
$$
Here $\times$ is matrix multiplication.
Let's now look at the information that takes 3 periods. This is the information that goes from $i$ to some intermediate $k$ then from $k$ to some intermediate $\ell$ and finally to $j$, so the path $i \to k \to \ell \to j$. The amount of information on this path is $W(i,k) \cdot W(k,\ell) \cdot W(\ell, j)$. If we sum over all these paths of length 3, we get:
$$
\sum_k \sum_\ell W(i,k) \cdot W(k,\ell) \cdot W(\ell, j) = (W \times W \times W)_{i,j} = (W^3)_{i,j}.
$$
If we continue, then we can generalize this to any path of length $T$. The information sent from $i$ to $j$ through all such paths is given by:
$$
(W \times W \times \ldots W)_{i,j} = (W^T)_{i,j}.
$$
Now if info goes from $i$ to $j$ either takes 1 period, 2 periods, 3 periods, ... $T$ periods to travel. As such, to compute the total expected pieces of info that $j$ receives from $i$ in $T$ periods is given by the sum over all these periods:
$$
H(W, T) = \sum_{t = 1}^T (W^t)_{i,j}.
$$
Now the total amount of info sent from $i$ and heard by someone can be obtained by taking the sum over all recipients $j$.
$$
DC(W,T) = \sum_{j} (H(W,T))_{i,j} = (H(W,T)\times 1)_{i,j}
$$
This amounts to adding up the rows of the matrix $H(W,T)$.
An example
Let's take the example of three individuals $1,2$ and $3$. Assume that $W$ is given by:
$$
W = \begin{bmatrix} 0 & 0.3 & 0.2\\ 0.2 & 0 & 0.6\\ 0 & 0.4 & 0\end{bmatrix}
$$
This implies that (for example) the probability that 2 hears from 1 in one period equals 0.3.
Then
$$
W^2 = \begin{bmatrix} 0.06 & 0.08 & 0.18\\ 0 & 0.3 & 0.04\\ 0.08 & 0 & 0.24\end{bmatrix}
$$
Here, for example, the amount received by 2 from 1 in two periods is $0.08 = 0.2 \times 0.4$
Next, multiplying $W^2$ once more by $W$ gives:
$$
W^3 = \begin{bmatrix} 0.016 & 0.09 & 0.06\\ 0.06 & 0.016 & 0.18\\ 0 & 0.12 & 0.016\end{bmatrix}
$$
Then:
$$
H(W, 3) = W + W^2 + W^3 = \begin{bmatrix} 0.076 & 0.47 & 0.44\\ 0.26 & 0.316 & 0.82\\ 0.08 & 0.52 & 0.256\end{bmatrix}
$$
So in three periods $2$ receives on average $0.47$ pieces of info from $1$
The total information sent from $1$ to someone is computed by taking the sum of the elements in row $1$, which gives:
$$
0.076 + 0.47 + 0.44 = 0.986
$$
In general:
$$
DC(W, 3) = \begin{bmatrix} 0.986\\1.396\\0.856\end{bmatrix}
$$
