The inequality is this:
$$
\label{eq:cut}
\overline{\beta}(\overline{\gamma} x - u) + h(y, v) \leq \overline{\beta}(\overline{\gamma} - t),
$$
valid on $y > 0$ when
$$
h(y,v) := A y + B g(y,v),\quad g(y,v) := \frac{yv}{y+v},\quad v \geq 0,
$$
where $A = \overline{\gamma}-\underline{\gamma} > 0$ and $B = \underline{\gamma} < 0$ are coefficients. The challenge is first to put this on conic quadratic form, and then to extend its validity to all of $y \in \mathbb{R}$, while staying tight on $y > 0$. On the first challenge of conic representability, our analysis led to a small surprise, namely, the possibility to represent harmonic means (see this blog post). The inequality $\eqref{eq:cut}$ is thus representable on $y > 0$ by $\overline{\beta}(\overline{\gamma} x - u) + h \leq \overline{\beta}(\overline{\gamma} - t)$ and
$$
\label{eq:repr:ypos}
h = A y + B g,\quad \left(\begin{matrix}0.5 (y-g)\\ y+v\\ y\end{matrix}\right) \in Q_r^3,\quad v \geq 0.
$$
We stress that the conic form is a proof of convexity. Next is the challenge of extending this representation to be valid on all of $y \in \mathbb{R}$, while staying tight on $y > 0$. That is, as proven in the paper, we need to extend the representation such that $h \leq 0$ on $y \leq 0$. Notably, however, the current representation in $\eqref{eq:repr:ypos}$ fails this criterion as shown by the graph of $h(y, 5)$ where $A=1$ and $B=-3$.
For this task we deploy another reformulation technique which, in lack of better names, may be denoted as stretching the extremum. The idea is to locally, in the representation of $h(y,v)$, exchange $y$ by $\hat{y} = y + s$ for some amount of stretch $\underline{s} \leq s \leq \overline{s}$. To see the effect of this let $y_{min} = \arg\min_{y \in \mathbb{R}} h(y,v)$, and note that
$$
\min_{s \in \mathbb{R}} h(y+s,v) = \begin{cases}
h(y+\overline{s},v), & {\rm if\ } y_{min} \geq y+\overline{s}, \\
h(y_{min},v), & {\rm if\ } y+\underline{s} \leq y_{min} \leq y+\overline{s},\\
h(y+\underline{s},v), & {\rm if\ } y+\underline{s} \geq y_{min}.
\end{cases}
$$
The visual interpretation of this is that we cut the graph of the function in two parts at $y_{min}$, move the two parts away from each other along the y-axis by amounts determined by $\underline{s}$ and $\overline{s}$, and finally fill in the gap with the extremum value $h(y_{min},v)$. In case of $\eqref{eq:repr:ypos}$ we use $\underline{s} = 0$ and $\overline{s} = \infty$ to allow $h$ to take smaller values of $h(y,v)$ at any higher value of $y$. In the graphed example of $h(y, 5)$ above, we get that the lower bound on $h$ is relaxed to the left of the extremum.
As $h(0,v) = 0$ for all $v \geq 0$, we get the desired property that $h \leq 0$ is possible on $y \leq 0$. In conclusion, the globally valid conic quadratic representation of the convex inequality we started out from, is given by $\overline{\beta}(\overline{\gamma} x - u) + h \leq \overline{\beta}(\overline{\gamma} - t)$ and
$$
h = A \hat{y} + B g,\quad \left(\begin{matrix}0.5 (\hat{y}-g)\\ \hat{y}+v\\ \hat{y}\end{matrix}\right) \in Q_r^3,\quad v \geq 0, \quad \hat{y} \geq y.
$$
