If you build the Taylor-McLaurin series at $x=-2$, you obtain $$2x^2-8=0 - 8 (x+2)+ 4 (x+2)^2/2 + O\left((x+2)^3\right)$$ Just retain the first two terms, zero and $-8(x + 2)$, which is a linear approximation of the original expression.

You get the same if you write $$2x^2-8=2\Big((x+2)-2\Big)^2-8=2\Big((x+2)^2-4(x+2)+4\Big)-8=-8(x+2)+2(x+2)^2$$

Let $u=x+2$ so that $u=0$ when $x=-2$ $$x^2=(u-2)^2$$ $$x^2=4-4u+u^2$$ $x^2$ is linerized as $x^2=4-4u=4-4(x+2)$ which is brought back into the ODE :

$$x'=2x^2-8=2(4-4(x+2))-8=8-8(x+2)-8$$ $$x'=-8x-16$$

$f(x) = 2 x^2-8$. We have $f(-2) = 0$ as expected. The linearized system is $y' = f'(-2) y$, which is $y'=-8y$.

If you want to express this as a system based around $-2$ rather than zero, let $z=y-2$, or $y=z+2$, which will give the equation: $z'=-8 z -16$.