Let's start from (EQ 5) (introduce $w$ notation for wealth factor and $C_i$ for call price).
$$ wC_i = \lambda_i (F_i -S_i)(S_{i+1}-S_i)^{-1} (S_{i+1}-s_i) +\Sigma $$
I have used (EQ 3) $p_i = (F_i -S_i)(S_{i+1}-S_i)^{-1}$.
This is equivalent to:
$$ wC_i (S_{i+1}-S_i)= \lambda_i (F_i -S_i) (S_{i+1}-s_i) +\Sigma (S_{i+1}-S_i)$$
$$ wC_iS_{i+1} -wC_iS_i=\lambda_iF_iS_{i+1}-\lambda_iF_is_i -\lambda_iS_i S_{i+1} +\lambda_iS_i s_i + \Sigma S_{i+1} - \Sigma S_i$$
$$ wC_iS_{i+1} -\lambda_iF_iS_{i+1} - \Sigma S_{i+1} = -\lambda_iF_is_i -\lambda_iS_i S_{i+1} +\lambda_iS_i s_i - \Sigma S_i +wC_iS_i $$
This is further equivalent to:
$$ S_{i+1}(\lambda_i F_i -wC_i +\Sigma)= \lambda_iF_is_i +\lambda_iS_i S_{i+1} -\lambda_iS_i s_i + \Sigma S_i -wC_iS_i $$
So the the denominator in (EQ 8) is correct and doesn't care about the extra relationship:
$$ S_iS_{i+1} = s_i^2.$$
We now use the extra relationship to process our right hand side:
$$ \lambda_iF_is_i +\lambda_iS_i S_{i+1} -\lambda_iS_i s_i + \Sigma S_i -wC_iS_i $$
$$ = \lambda_iF_is_i +\lambda_is_i^2 -\lambda_iS_i s_i + \Sigma S_i -wC_iS_i $$
$$ = (\lambda_iws_i^2 + \lambda_is_i^2) - (wC_iS_i + \lambda_iS_i s_i - \Sigma S_i)$$
(I used $F_i =w s_i$ in the last equality.)
The second parenthesis pair certainly resembles the numerator in (EQ 8), but I have no idea why subtracting it from the first parenthesis pair would produce the coveted numerator.
Any help from community (or different approach/answer) is more than welcome.