$[x^2]$ jumps at integer values of $x^2$ within $(0,4)$, i.e., $x^2=1,2,3$, giving $x=\pm 1,\pm \sqrt{2},\pm \sqrt{3}$.

At $x=0$, $|x|\cdot [x^2]=0$ and the limit is $0$, so $g$ is continuous there. Checking the six points above, $g$ is discontinuous at all of them. So $S=\{-1,1,-\sqrt{2},\sqrt{2},-\sqrt{3},\sqrt{3}\}$.

For $f(x)=\min (\sqrt{2}x,x^2)$: when $x<0$, $\sqrt{2}x<0<x^2$, so $f(x)=\sqrt{2}x$. When $x>0$, $f(x)=\sqrt{2}x$ if $\sqrt{2}x\le x^2$ (i.e., $x\ge \sqrt{2}$), else $x^2$.

Evaluating: $f(-1)=-\sqrt{2}$, $f(1)=1$, $f(-\sqrt{2})=-2$, $f(\sqrt{2})=2$, $f(-\sqrt{3})=-\sqrt{6}$, $f(\sqrt{3})=\sqrt{6}$.

Sum $=-\sqrt{2}+1-2+2-\sqrt{6}+\sqrt{6}=1-\sqrt{2}$.