In this paper, on page 243, we have
$$d_{jk} = \frac{a_j}{a_k(x_j - x_k)}$$
where $$a_k = \prod_{l = 0;l\neq k}^{N}(x_k-x_l)$$
Now, $a_k$ requires evaluating N multiplications. Why does the author say, "$N^2$ operations are required to compute $a_j$"?
Of course you are right, $a_k$ requires N multiplications, it's obvious. It does not involve $d_{jk}$...
[EDITED because of my inattention] I suspect "operations" here means multiplies and inverses only, neglecting subtractions $x_j-x_k$.
Taking a look at the paper, I see Eq. (2.8):
$$ \phi_{N+1}(x) = \prod_{l=0}^N (x-x_l)\,. $$ Then, [Eq. (2.9)] $$ \phi'_{N+1}(x) = \sum_{k=0}^{N}\prod_{l=0,l\neq k}^N (x-x_l) $$ and so [Eq. (2.10)] $$ \phi'_{N+1}(x_k) = a_k \,.$$
For each $a_k$, $N$ multiplies are required. Could one interpret the assertion globally, on all $a_j$? Indeed, the paper evaluate $4N^2$ operations for the whole matrix $D$. What if one only counts multiplies or inverses? Back in 1989, it was not uncommon to forget adds and subtractions in complexity. In Eq. 2.12a, they evaluate off-diagonal elements: $$\frac{a_j}{a_k(x_j-x_k)}$$ with "another $2N^2$ operations". Once the $a_j$ are computed, this seems sound: $1$ product and $1$ divide for each pair of $(j,k)$, $j\neq k$. Then they count $N^2$ for the diagonal elements $$\sum_{l=0,l\neq k}^{N}\frac{1}{x_k-x_l}\,,$$ ie $N$ inverses for each of the diagonal elements, about $N^2$ total.
