Thank for your answer. I am interested in your last comment. You suggested that $\mathbb{R}((X))$ can be regarded as a subfield of hyperreal field by a non-canonical way. I would like a detailed explanation of "non-canonical way".

Sorry. I did not explain the strong transitivity. Let $\mathcal{A}$ be a system of apartments, now in this case, $\mathcal{A}=\{gA\mathrel{\vert}g\in G\}$. We say that the action of $G$ is strongly transitive if for any $A_{1},A_{2}\in \mathcal{A}$ and chambers $C_{1}\subset A_{1}, C_{2}\subset A_{2}$, there exists $g\in G$ such that $g(A_{1},C_{1})=(A_{2},C_{2})$.

Sorry. The term of "remaining component" had no special meaning. I understand that $\langle S\rangle \neq \langle s_{1},s_{2} \rangle\cup \langle w_{1}\rangle$.

Now I proved a part of 2. The intersection $\phi(N)\cap \hat{B}$ is normal in $\phi(N)$. For any $x,y\in N$ such that $\phi(x)\in \hat{B}$, we have $\phi(xBx^{-1})=\phi(B)$ from the definition ${\rm Stab} B$. So we have $\phi(xb_{1}x^{-1}b_{2})=0$ for any $b_{1}\in B$ and some $b_{2}\in B$. Since ${\rm Ker}\phi\subset B$, $xb_{1}x^{-1}b_{2}\in B$. As $N_{G}(B)=B$, this means that $x\in B\cap N (\triangleleft N)$. Then we have $\phi(yxy^{-1})\in \hat{B}\cap \phi(N)$. Thus $\hat{B}\cap \phi(N)\triangleleft \phi(N)$.

I can prove 1. For any $g\in \hat{G}$, there exists $h\in G$ such that $\phi(h^{-1})g\in {\rm Stab} B$. Since $\phi(G)\subset \hat{G}_{0}$, we have $\hat{G}=\hat{G}_{0}\cdot {\rm Stab} B$. Thus the restriction of $\xi$ to ${\rm Stab} B$ is surjective.

Thank you. It's a careless mistake.