= Solution
Write $m_j=\mathbb EY^j$ and $M=M_Y$. The derivatives of the <cumulant-generating function> are
$$
\kappa_Y'=\frac{M'}M,\qquad \kappa_Y''=\frac{M''}M-\frac{(M')^2}{M^2},\qquad \kappa_Y'''=\frac{M'''}M-\frac{3M'M''}{M^2}+\frac{2(M')^3}{M^3}.
$$
Since $M(0)=1$ and $M^{(j)}(0)=m_j$, the first three <cumulants> are
$$
\boxed{\kappa_{Y,1}=m_1=\mathbb EY,\quad \kappa_{Y,2}=m_2-m_1^2=\operatorname{Var}(Y),\quad \kappa_{Y,3}=m_3-3m_1m_2+2m_1^3=\mathbb E(Y-\mathbb EY)^3.}
$$
The last equality follows by expanding the cube and taking <expectations>. We use the paper's unstandardized third <central moment> throughout. The usual <skewness> coefficient divides this quantity by $\operatorname{Var}(Y)^{3/2}$; its sign is the same whenever the <variance> is positive.
There is a minor domain qualification: finite ordinary <moments> do not guarantee a finite <moment-generating function> for positive arguments. The differentiations are valid in a neighborhood of zero when a positive exponential moment exists. For positive variables with the stated finite <moments>, all formulas also follow from left derivatives at zero, because $Y^j e^{tY}\le Y^j$ for $t\le0$ and <dominated convergence> applies. Thus no extra exponential-moment assumption is needed for the moment identities themselves.
For the <random sum of independent claims>, condition on $N=k$. Independence of the claims and independence from the count give
$$
\mathbb E[e^{tS}\mid N=k]=M_{X_1}(t)^k.
$$
Taking <expectations> and then logarithms yields the <random-sum transform identity> and the <cumulant composition for a random sum>:
$$
\boxed{M_S(t)=G_N(M_{X_1}(t)),\qquad\kappa_S(t)=\kappa_N(\log M_{X_1}(t))=\kappa_N(\kappa_{X_1}(t)).}
$$
Here $G_N$ is the <probability generating function>. These identities hold wherever the transforms are finite, in particular for $t\le0$ with positive claims.
For reuse in the three cases, write $m_j=\mathbb EX_1^j$ and $a_j=\kappa_{N,j}$. Applying the chain rule three times gives
$$
\kappa_{S,3}=a_1(m_3-3m_1m_2+2m_1^3)+3a_2m_1(m_2-m_1^2)+a_3m_1^3,
$$
or equivalently
$$
\kappa_{S,3}=a_1m_3+3(a_2-a_1)m_1m_2+(a_3-3a_2+2a_1)m_1^3.
$$
This expresses the requested third <central moment> in raw claim <moments>, while keeping the contribution of claim-count variability explicit.
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