In the proof of the Lemma 5.2, I was puzzled by the statement "$v$ generates a finite dimensional factor $F_1$ and...".
Why can $v$ generate a finite dimensional? If $v^i\neq v^j$ for all $i\neq j$, then the von Neumann algebra generated by $v$ is infinite dimensional.
How to check $\varphi_{F_2}=\omega_{\lambda_2}$? where $\omega_{\lambda_2}(e_{11})=\lambda_2, \omega_{\lambda_2}(e_{22})=\frac{ \lambda_2}{1+\lambda_2},\omega_{\lambda_2}(e_{12})=\omega_{\lambda_2}(e_{22})=0$ and $e_{11},e_{12},e_{21},e_{22}$ are matrix units of $M_2(\Bbb C)$.
Not if $v^*v+vv^*=1$. This forces the two projections $v^*v$ and $vv^*$ to be pairwise orthogonal. Then $$ v^2=v(v^*v)(vv^*)v=0. $$ The four elements $v^*v, v, v^*, vv^*$ behave exactly like the matrix units in $M_2(\mathbb C)$. Hence the von Neumann they generate is (a copy of) $M_2(\mathbb C)$.
I don't know the theory the authors are using, so I will just comment on the arithmetic. If you find a partial isometry $v$ with $v\varphi=\lambda_2\varphi v$ and $v^*v+vv^*=1$, then you have $$ \varphi(vv^*)=\lambda_2\varphi(v^*v),\qquad\qquad \varphi(v^*v)+\varphi(vv^*)=1. $$ Solving the $2\times 2$ linear system gives you $$ \varphi(v^*v)=\frac{\lambda_2}{1+\lambda_2},\qquad\varphi(vv^*)=\frac1{1+\lambda_2}. $$ You also have $\varphi(v)=\lambda_2\varphi(v)$, so as long as $\lambda_2\ne1$ you get $\varphi(v)=0$.