Why is 2H2O2 -> 2H2O + O2 a first order reaction?
Summarizing from the excellent comments:


  Isn't the order of the reaction supposed to be the sum of powers of the concentration of the reactants?


Yes, that is how you get the overall order of reaction. If you compare the rate with a given set of concentrations to the rate with all concentrations doubled, the rate goes up by two to the power of the overall order.


  I don't see why that would be first order?


It is not, but it is pseudo-first order. If you start the reaction using certain initial concentrations, you will find an exponential decay of the reaction rate over time (i.e. there is a constant half-life).

As JM97 comments, the iodide is a catalyst, so its concentration does not change over time. So while the concentration of hydrogen peroxide drops over time and causes the reaction to slow down, the iodide concentration in the rate law does not contribute to a slowing down (no matter what order the reaction is in iodide).


  [OP responding to Ivan...] Actually the textbook statement is "_This reaction is first order with respect to both H_2 O_2 and I^- "


As Ivan comments, that means the reaction rate is proportional to the first power of the hydrogen peroxide concentration (if the iodide concentration is unchanged). It is also proportional to the first power of the iodide concentration (if the hydrogen peroxide concentration is unchanged). If both are changed at the same time, you use the rate law to see what the effect is.