Effect of concentration on rate of reaction
The fast reaction should be written as reversible reaction if $\ce{NOBr2}$ is a high energy intermediate:
$$\ce{NO + Br2 &lt;=&gt; NOBr2}$$

Then, the concentration of $\ce{NOBr2}$ can be estimated as:

$$[\ce{NOBr2}] = K_{eq} [\ce{Br2}] [\ce{NO}]$$

If the second reaction is an elementary step, the rate law follows the stoichiometry:

$$\text{rate} = k [\ce{NOBr2}] [\ce{NO}]$$

Substituting the concentration of the intermediate $\ce{NOBr2}$ into that rate law gives the overall rate law:

$$\text{rate} = k K_{eq} [\ce{Br2}] [\ce{NO}] [\ce{NO}]$$

and the dependency of the rate on doubling all reactants.


  Now I know that overall order of a reaction is equal to the molecularity of the slowest step


This is only true if the slowest step is the first step. Otherwise, you have to consider all the reactants that make the intermediates necessary for the slow step.

What if the first reaction goes to completion?
In this case, the concentration of the intermediate would depend on the limiting reactant. Changing all reactant concentration by a factor of two, the concentration of intermediate would increase by a factor of two as well (not like the factor of 4 if the first step is a pre-equilibrium).

What else do we need to know to solve this problem?
The problem description does not tell us the physical state of the reactants, intermediates and products. If $\ce{Br2}$ is a pure liquid, we would not be able to change its concentration. I would guess the intention is to have all species in the gas phase, though.