What is the order of the reaction with respect to NO?
The rate of that second reaction will indeed be first order with respect to $\ce{NO}$:

$$
R = R_2 = k_2[\ce{NO}][\ce{NOBr2}].
$$

The problem is, the overall order of the reaction is determined from the rate expression involving only the reactants.  The above rate expression is written in terms of $\ce{NO}$ (reactant) and $\ce{NOBr2}$ (intermediate), not in terms only of reactants.

To convert the reaction rate expression to one involving only reactants, exploit the fast equilibrium of the first reaction,

$$
K_1 = \frac{[\ce{NOBr2}]}{[\ce{NO}][\ce{Br2}]},
$$

to find an expression for $[\ce{NOBr2}]$ in terms only of $K_1$ and the reactants:

$$
[\ce{NOBr2}] = K_1[\ce{NO}][\ce{Br2}].
$$

The expression for $R = R_2$ can thus be rewritten as

$$
R=R_2 = k_2[\ce{NO}]\left(K_1[\ce{NO}][\ce{Br2}]\right) = k_2K_1[\ce{NO}]^2[\ce{Br2}],
$$

which is indeed 2$^{\mathrm{nd}}$ order in $\ce{NO}$.