This is not answer in general to your question but a rather nice confirmation in the case of $\vec{\Gamma} = B_n $. The dynkin diagram $B_n$ has two leaves --- one joined by a type $A_2$-bond (weight $3$ by your convention) and another joined by a type $B_2$-bond (weight $4$ by your convention). So there there ought to be two distinct semi-direct product factorisations of the unipotent radical $N_{B_n}$ according to your conjecture.
As I noted in your other posting the Kac-Moody group $G_{B_n}$ in this case is the symplectic group $\text{Sp}_{2n} \big( \Bbb{C} \big)$ whose unipotent radical $N_{B_n}$ can be identified as the group of all $2n \times 2n$ invertible complex matrices having the following block-decomposition
\begin{equation} \left( \begin{array}{c|c} A & A \dot{w} S \\ \hline \\ \Bbb{O} & \dot{w}A^{-T} \dot{w} \\ \end{array} \right) \end{equation}
where $A$ is any $n \times n$ unipotent matrix (i.e. upper-triangular with $1$'s on the diagonal), $S$ is any $n \times n$ symmetric matrix, and $\dot{w}$ is the $n \times n$ permutation matrix of $w \in S_n$ defined by $w(i) = n+1 -i$ for $1 \leq i \leq n$.
Consider first the case of the leaf attached by the $B_2$-bond. Let $H$ denote the abelian (indeed additive) subgroup of $N_{B_n}$ consisting of matrices of the form
\begin{equation} \left( \begin{array}{c|c} \Bbb{I} & \dot{w} S \\ \hline \\ \Bbb{O} & \Bbb{I} \\ \end{array} \right) \end{equation}
It's easy to see that $H$ is normal in $N_{B_n}$ and as a abelian group isomorphic to the vector space $\mathcal{H}$ consisting of all symmetric complex matrices. The quotient $N_{B_n} \big/ H$ is isomorphic to the group $N_{A_{n-1}}$ of all $n \times n$ unipotent complex matrices which acts linearly on $\mathcal{H}$ by
\begin{equation} A \cdot S \ = \ A \,\dot{w} \, S \, \dot{w} \, A^T \end{equation}
and consequently we have the semi-direct product factorisation $N_{B_n} \cong \mathcal{H} \rtimes N_{A_{n-1}}$.
Consider the leaf with the type $A_2$-bond. In view of your conjecture we should expect a semi-direct factorisation involving $N_{B_{n-1}}$ instead of $N_{A_{n-1}}$. To effect this decomposition let's embed $\text{Sp}_{2n-2}\big( \Bbb{C} \big)$ into the central $(2n-2) \times (2n-2)$ principle block of $\text{Sp}_{2n} \big( \Bbb{C} \big)$, namely
... to be continued at 5pm