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 \ = \ \dot{w} \, A \,\dot{w} \, S \, \dot{w} \, A^{T} \, \dot{w}
\end{equation}

and consequently we have the semi-direct product factorisation 
$N_{B_n} \cong \mathcal{H} \rtimes N_{A_{n-1}}$ expressed as

\begin{equation} \big(S_1,A_1\big) \circ \big(S_2,A_2 \big) \ = \ \Big(S_2 + A_2^{-1} \cdot S_1 \, , \,  A_1A_2 \Big) \end{equation}

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 $(2n-2) \times (2n-2)$ block of $\text{Sp}_{2n} \big( \Bbb{C} \big)$, namely

\begin{equation}
\left(
\begin{array}{c|c|c}
1 &  &   \\ 
\hline 
 & \text{Sp}_{2n-2}\big( \Bbb{C} \big) &  \\
\hline 
 &  & 1
\end{array}
\right)
\end{equation}

Let $\acute{H}$ be the subgroup of $N_{B_n}$ consisting of matrices
having the block-form

\begin{equation}
\left(
\begin{array}{c|c}
U & V  \\ 
\hline  
\Bbb{O} & \dot{w}U^{-T}\dot{w} \\
\end{array}
\right)
\end{equation}

where 

\begin{equation}
U \ = \ \left( 
\begin{array}{c|l}
1 & \vec{u}^{ \ \scriptscriptstyle T}  \\ 
\hline  
\vec{0} & \Bbb{I} \\
\end{array}
\right) \quad \text{and} \quad 
V \ = \ \left(
\begin{array}{c|l} \vec{v}^{\ \scriptscriptstyle T} & v  \\ 
\hline  
\Bbb{O} & \vec{v} \\
\end{array}
\right)
\end{equation}

for any vectors $\vec{u}$ and $\vec{v}$ in $\Bbb{C}^{n-1}$ and any
$v \in \Bbb{C}$. Again $\acute{H}$ is a normal, it trivially intersects
$N_{B_{n-1}}$ with respect to the embedding of $\text{Sp}_{2n-2}\big( \Bbb{C} \big)$ in $\text{Sp}_{2n}\big( \Bbb{C} \big)$, and $N_{B_{n-1}} \, \acute{H}
= N_{B_n}$. Let $\acute{\mathcal{H}}$ denote $\acute{H}$ when viewed as a vector space; the conjugation action of $N_{B_{n-1}}$ induces a linear
action of $N_{B_{n-1}}$ on $\acute{\mathcal{H}}$ and so we get a second
semi-direct product factorisation $N_{B_n} \cong \acute{\mathcal{H}} \rtimes 
N_{B_{n-1}}$ as anticipated.

... to be continued at 5pm