Given,

Functional equation $f\left(x+y\right)=f\left(x\right)+f\left(y\right)-1$

Now taking $x=0 \& y=0$ in above equation we get,

$f\left(0+0\right)=f\left(0\right)+f\left(0\right)-1\Rightarrow f\left(0\right)=1$

Now we know that,

$f^{\text{'}}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

$\Rightarrow f^{\text{'}}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x\right)+f\left(h\right)-1-f\left(x\right)}{h}$

$\Rightarrow f^{\text{'}}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(h\right)-1}{h}$

Now let $\underset{h\to 0}{\lim }\frac{f\left(h\right)-1}{h}=k$

So, the equation becomes $f^{\text{'}}\left(x\right)=k$

Now putting $x=0$ in above equation we get,

$f^{\text{'}}\left(0\right)=k\Rightarrow k=2$ $\left\{\text{as given }{f}^{\text{'}}\left(0\right)=2\right\}$

So, $f^{\text{'}}\left(x\right)=2$

Now integrate both side we get,

$f\left(x\right)=2x+c$

Now again taking $x=0$ we get,

$f\left(0\right)=2\times 0+c\Rightarrow c=1 \left\{\text{as }f\left(0\right)=1\right\}$

So, $f\left(x\right)=2x+1$

And $f\left(-2\right)=2\times \left(-2\right)+1\Rightarrow f\left(-2\right)=-3$

Hence, $\left|f\left(-2\right)\right|=3$