Steady-state approach - how to find rate equation for intermediate?
Split the scheme into three parts and write down each step. Start by definition with
$$\displaystyle -\frac{1}{2}\frac{d[\ce{NO}]}{dt}=+\frac{d[\ce{N2O2}]}{dt}$$
then the first step $\displaystyle -\frac{1}{2}\frac{d[\ce{NO}]}{dt}=k_1[\ce{NO}]^2\quad$ or $\quad\displaystyle \frac{d[\ce{NO}]}{dt}=-2k_1[\ce{NO}]^2$
and so $\ce{N_2O_2}$ is produced as
$$\displaystyle \frac{d[\ce{N2O2}]}{dt} = k_1[\ce{NO}]^2$$
In the reverse equilibrium step $\ce{N2O2}$ is lost (by dissociatiing) so $\displaystyle \frac{d[\ce{N2O2}]}{dt} = -k_1^{'}[\ce{N2O2}]$
and in the last reaction it is also lost by reaction with oxygen therefore $\displaystyle \frac{d[\ce{N2O2}]}{dt}=-k_2[\ce{N2O2}][\ce{O2}]$.
Adding up all three rates for $\ce{N_2O_2}$ gives
$$\displaystyle \frac{d[\ce{N2O2}]}{dt}=k_1[\ce{NO}]^2-k_1^{'}[\ce{N2O2}]-k_2[\ce{N2O2}][\ce{O2}]$$
You can now find the steady-state conditions.
(In your last equation there is $2k_1$ and it is quite common to see it written this way, all it does is to change $k_1$.)