jmc

Define a new function $h(x)$ so that $h(x)=g(f(x))$. Then $$\begin{array}{rl}h(x)& =g\left(\frac{1+x}{1-x}\right)=\frac{-2}{\frac{1+x}{1-x}+1}\\ & =\frac{-2}{\frac{1+x}{1-x}+\frac{1-x}{1-x}}=\frac{-2}{\frac{2}{1-x}}\\ & =\frac{-1}{\frac{1}{1-x}}=-(1-x)=x-1.\end{array}$$Thus our desired value is equivalent to $8$ compositions of the function $h(x)$. For each composition we subtract $1$ from the input value, so for $8$ compositions we subtract a total of $8$ from the input value, which is $12$. So our answer is $12-8=\boxed{4}$.