[guided]In the standard model, the blow-up is not merely the operation of deleting a ball; the boundary of the deleted ball is collapsed along a circle foliation. The relevant circle action is the diagonal action
\begin{align*}
e^{i\theta}\cdot(z_1,z_2)=(e^{i\theta}z_1,e^{i\theta}z_2).
\end{align*}
Its Hamiltonian, in the capacity normalization used in the theorem, is the map
\begin{align*}
H:\mathbb C^2\to\mathbb R,\qquad H(z_1,z_2)=\pi(|z_1|^2+|z_2|^2).
\end{align*}
The ball of capacity $\varepsilon$ is precisely the sublevel set $H<\varepsilon$, and its boundary is the level set $H=\varepsilon$.
We now invoke the symplectic cut construction [citetheorem:10074]. Its hypotheses are satisfied in this local model: the diagonal $S^1$-action is Hamiltonian with moment map $H$, and on the sphere $H^{-1}(\varepsilon)$ the action is free because a point on this sphere is not $(0,0)$, so at least one coordinate is nonzero and a diagonal phase fixing the point must be the identity. The construction removes the open part $H<\varepsilon$ and replaces the boundary level by the quotient $H^{-1}(\varepsilon)/S^1$, which is the exceptional divisor of the blow-up.
Now translate this operation into the moment image. Since
\begin{align*}
\mu_{\mathbb C^2}(z_1,z_2)=(\pi |z_1|^2,\pi |z_2|^2),
\end{align*}
the equation $H(z_1,z_2)=\varepsilon$ becomes
\begin{align*}
x_1+x_2=\varepsilon.
\end{align*}
The inequalities $x_1\ge 0$ and $x_2\ge 0$ remain because each $x_i$ is a squared norm multiplied by $\pi$. Therefore the collapsed boundary contributes the new closed segment
\begin{align*}
\{(x_1,x_2)\in\mathbb R^2:x_1\ge 0,\ x_2\ge 0,\ x_1+x_2=\varepsilon\}.
\end{align*}
The open triangle
\begin{align*}
\{(x_1,x_2)\in\mathbb R^2:x_1\ge 0,\ x_2\ge 0,\ x_1+x_2<\varepsilon\}
\end{align*}
is removed. Hence the post-blow-up local moment image is
\begin{align*}
\{(x_1,x_2)\in\mathbb R^2:x_1\ge 0,\ x_2\ge 0,\ x_1+x_2\ge\varepsilon\}.
\end{align*}
The distinction between the strict inequality for the removed ball and the non-strict inequality for the resulting polygon is exactly the boundary convention in the statement.[/guided]