Step 1: Go back to the rate law obtained from the slow step.
From the mechanism, $\text{Rate} = k[H_2O_2]^1[I^-]^1$. The order with respect to each species is simply the power it is raised to in this expression, not anything guessed from the overall balanced equation.
Step 2: Read off the individual orders.
$H_2O_2$ appears to the power 1, so the reaction is first order in $H_2O_2$. $I^-$ also appears to the power 1, so the reaction is first order in $I^-$ as well.
Step 3: Add them up for the overall order.
The overall order of a reaction is just the sum of all the individual orders, so here it comes to $1 + 1 = 2$.
\[ \boxed{\text{Order in } H_2O_2 = 1,\ \text{Order in } I^- = 1,\ \text{Overall order} = 2} \]