The KKS formula is independent of the Lie-algebra representatives

Put $\zeta :={\xi }^{\mathrm{\prime }}-\xi$. The hypothesis gives ${\zeta }_{\mathcal{O}}(\beta )=0$, so $\zeta$ lies in the kernel of the infinitesimal orbit map, that is $\zeta \in {\mathfrak{g}}_{\beta }$ by [F2].

For every $\eta \in \mathfrak{g}$, [F1] evaluates the vanishing field at $\eta$ as ${\zeta }_{{\mathfrak{g}}^{\ast }}(\beta )(\eta )=\beta ([\zeta ,\eta ])=0$. Hence $\beta ([{\xi }^{\mathrm{\prime }},\eta ])=\beta ([\xi ,\eta ])+\beta ([\zeta ,\eta ])=\beta ([\xi ,\eta ])$ by bilinearity of the bracket.

The second argument is treated by alternation: if ${\eta }_{\mathcal{O}}(\beta )={\eta }_{\mathcal{O}}^{\mathrm{\prime }}(\beta )$, then $\beta ([\xi ,{\eta }^{\mathrm{\prime }}])=\beta ([\xi ,\eta ])$ by applying step 2.1 to $\eta ,{\eta }^{\mathrm{\prime }}$ and using $\beta ([\zeta ,\cdot ])=0$ for $\zeta \in {\mathfrak{g}}_{\beta }$; equivalently, the form $\beta ([\cdot ,\cdot ])$ is alternating, so its value depends skew-symmetrically on the two arguments.

Since every tangent vector of $\mathcal{O}$ at $\beta$ equals ${\xi }_{\mathcal{O}}(\beta )$ for some $\xi \in \mathfrak{g}$ by [F2], steps 2.1 and 3.1 show that the KKS prescription depends only on the two tangent vectors, so it defines a unique bilinear alternating form on $T_{\beta }\mathcal{O}$.