Why CLRS example on residual networks does not follows its formula?
That's not what the formula gives you.  As the caption says, the capacity of the augmenting path in the residual network in (b) is $4$.  Therefore we send 4 units of flow along the augmenting path from $s$ to $t$, namely, the path $s \to v_2 \to v_3 \to t$.  In particular, $f(s,v_2)=8$, $f'(s,v_2)=4$, and $f'(v_2,s)=0$, so the updated flow is $8+4-0=12$.