I understand how to use $y = e^{rx}$ for solving homogeneous 2 order differential equation, but can't use this method somehow for $y'' = 0$.
$y'' = r^2e^{rx}$
$r^2e^{rx} = 0$
$r^2 = 0 => r = 0$
General solution of 2 order diff is $y = C_1e^{r_1x} + C_2e^{r_2x}$ We only got one $r$ which is 0, so $y = C_1e^{0x}$ => $y = C_1$
How do I get $y = c_1x + c2$ by this approach ?
Your solution $r=0$ is a twice repeated root, so the general solution is actually $$y=(c_1x+c_2)e^{rx}=(c_1x+c_2)e^{0x}=c_1x+c_2$$
You can see this normally since a (homogeneous) linear differential equation of order $2$ must have $2$ linearly independent solutions. You found one solution $y_1(x)=C_1e^{rx}=C_1$. You should think of a second solution $y_2(x)$, linearly independent from $y_1(x)$ i.e. not a constant multiple of $y_1(x)$, that you will guess. The simplest function linearly independent of $y_1(x)$ is $y_2(x)=C_2xe^{rx}$. Then $$y_2''=C_2re^{rx}(rx+2)=0$$ since $r=0$, so $y_2(x)=C_2xe^{rx}=C_2x$ is another solution. Since we can have a maximum of two linearly independent solutions to a second order linear differential equation, that tells us $y_g(x)=C_1+C_2x =c_1x+c_2$, with $C_1=c_2$ and $C_2=c_1$.