Two corrections to the published construction. Both are recorded here because the construction below is the corrected one.

The first is the number of segments. The paper takes $R = k(n-1)+1$, and that is one too few. The extraction of a hitting set is a pigeonhole: each machine's offset is nondecreasing across segments and confined to $\{1, \dots, n\}$, so it changes at most $n-1$ times, and at most $k(n-1)$ of the gaps between consecutive segments see any change. A gap across which nothing changes therefore exists only if there are more than $k(n-1)$ gaps, that is only if $R - 1 > k(n-1)$. The published value gives exactly $R - 1 = k(n-1)$, and the count is not slack: let the first machine change across the first $n-1$ gaps, the second across the next $n-1$, and so on, and every gap is consumed with no two segments alike. Taking $R = k(n-1)+2$ repairs it, and costs nothing: every statement about the construction is generic in $R$.

The second is the range of $k$. The paper writes $p(A^r_{i,j}) = \frac{1}{k-1} g(r,j) Q$, which presupposes $k \ge 2$ and says so nowhere. At $k \le 1$ the construction does not merely degenerate, it is wrong: there $Q = 0$, so every second operation has length zero, every interval $[s, d)$ is empty and no two jobs conflict at all. Every dummy still costs more to preprocess than it has time for, but all selection jobs become schedulable simultaneously, whatever the family looks like, and the target is met with no hitting set in sight. The problem reduced from is accordingly restricted to $2 \le k \le n$, which costs nothing.

Two smaller things: the printed $\frac{1}{k-1} g(r,j) Q$ is the integer $g(r,j)(n+1)$, which is what appears below, and the target size printed in the opening paragraph of the section as $2m(2k-1)$ is the $R \cdot m \cdot (2k-1)$ that the section's own subsections and lemmas use.

Numbering. Jobs are numbered rather than tagged, because an instance is something a machine is handed and a word presents its jobs in an order. The selection block comes first, $R$ copies of one job per membership pair $(j, i)$ with $i \in F_j$, in the order in which the pairs are enumerated; then the two dummy blocks, each $R \cdot m \cdot n$ jobs indexed by segment, set and element in that order. Only the memberships that exist are given jobs, so the weights are all one and the hardness claim lands on the unit-weight slice, which is the problem $FF(1,m) \mid\mid \sum_j Z_j$ the paper's first theorem is about.

The universe is counted from zero here and from one in the paper, and the offset that distinguishes the $n$ jobs of an epoch from one another is correspondingly $i+1$. It has to be nonzero: an offset of zero would make two jobs of one epoch share a due date, and the extraction of the hitting set reads the element off the due date.

The element offsets are smaller than $Q$, so they perturb the due dates without disturbing the tiling of an epoch. That $n < Q$ is exactly what $k \ge 2$ buys.