$\DeclareMathOperator{\Nef}{Nef}$There is more to the situation in the paper than you have written here. I will use the notation of the paper instead of your notation.
In the paper, $X$ is a rational elliptic surface. We write $F$ for the class of a fiber, and $E$ signifies a $(-1)$-curve. The nef cone $\Nef X$ is spanned by these classes (as $E$ varies over all the $(-1)$-curves). These induce curve classes $F_{[n]}$ and $E_{[n]}$ on $X^{[n]}$. There is also the class $C_0$ of a curve contracted by the Hilbert-Chow morphism.
The cone $\Lambda \subset N^1(X^{[n]})$ is the cone of divisors which are positive on the curve classes $F_{[n]}$, $C_0$, and $E_{[n]}$. It contains the nef cone $\Nef(X)$ embedded in $\Nef(X^{[n]})$.
Kopper then considers the cone $\Lambda' \subset N^1(X^{[n]})$ which is spanned by $\Nef(X)$ and $(n-1)F^{[n]}-\frac{1}{2}B$, and proves that $\Lambda \subset \Lambda'$.
Crucially, Kopper computes at the end of the proof of Lemma 3.1 that the curve classes $E_{[n]}$ and $C_0$ are nonnegative on $(n-1)F^{[n]}-\frac{1}{2}B$, and therefore nonnegative everywhere on $\Lambda'$. (This remark at the end of Lemma 3.1 perhaps seems slightly out of place, and perhaps should have been discussed in the paragraph following the proof.) Thus, $\Lambda\subset \Lambda'$ can be cut out by the single additional restriction that the intersection with the curve class $F_{[n]}$ has to be nonnegative.
The picture is that $\Lambda$ is just the intersection of the cone $\Lambda'$ with the half-space determined by the inequality $D.F_{[n]}\leq 0$. The extremal rays of $\Lambda'$ are all either in $\Nef(X)$ or the single ray $(n-1)F^{[n]}-\frac{1}{2}B$. The only places $\Lambda$ can have new extremal rays is along the plane $D.F_{[n]}=0$, since the cone doesn't change anywhere else. And the hyperplane section of $\Lambda'$ gotten by intersecting with $D.F_{[n]}=0$ basically looks like a copy of $\Nef(X)$, since $\Lambda'$ is a cone over $\Nef(X)$. The extremal rays all lie on the lines between extremal rays of $\Nef(X)$ and the divisor $(n-1)F^{[n]}-\frac{1}{2}B$.
Also see the paper https://arxiv.org/abs/1509.04722 where a slightly easier situation is considered by similar methods in Section 5.