The particles have mass $m=8$ and we assume that particles $i$ and $j$, having coordinates $x_i$ and $x_j$, interact via the potential $$ U_{ij}(r_{ij})=16+q_iq_j\frac{16}{\cosh(r_{ij})},~r_{ij}=x_j-x_i. $$
The Hamiltonian of the three-particle system is $$ H=\frac{m}{2}\sum_{j=1}^3 v_j^2+U_{12}(r_{12})+U_{13}(r_{13})+U_{23}(r_{23})~~(*), $$ where $v_j=dx_j/dt$.
Without loss of generality, we assume that the total momentum in the system is equal to zero, i.e. $m(v_1+v_2+v_3)=0$.
Introducing new variables $$ x_2-x_1\to\sqrt{3}\alpha+\beta,~~x_3-x_1\to 2\beta,~~t\to\sqrt{2m}t, $$ the Hamiltonian of $(*)$ can be presented in the form $$ H=\frac{1}{2}\left(\dot{\alpha}^2+\dot{\beta}^2\right)+U_{12}(\sqrt{3}\alpha+\beta)+U_{13}(2\beta)+U_{23}(\sqrt{3}\alpha-\beta)~~(2), $$ which is the Hamiltonian of a unit-mass particle moving in the two-dimensional scattering potential.
I have two-questions to the yellow box content:
(1) How do I get from the Hamiltonian $(*)$ to the new Hamiltonian $(2)$?
(2) Why is $(2)$ the Hamiltonian of a unit-mass particle moving in the two-dim. scattering potential? I do not understand how this is meant.