Given $f\left(x\right)=\sqrt{2-x^2}$ and $g\left(x\right)=\ln \left(1-x\right)$

$\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)=\sqrt{2-x^2}+\ln \left(1-x\right)$

Let domain of $f\left(x\right)=\sqrt{2-x^2}$ be $D_1$,

then $2-x^2\ge 0\Rightarrow x^2-2\le 0$

i.e. $D_1\equiv \left[-\sqrt{2},\sqrt{2}\right]$

Let domain of $g\left(x\right)=\ln \left(1-x\right)$ be $D_2$

then $1-x>0\Rightarrow x-1<0$

i.e. $D_2\equiv \left(-\infty ,1\right)$

Hence, domain of $f\left(x\right)+g\left(x\right)$ will be $D_1\cap D_2\equiv [-\sqrt{2},1)$