Why does the rate of reaction increase less than proportionally to the increase in mass of catalyst?
There is a lot going on here. Part of the problem is poor experimental design. I'll just throw this out without evaluating all the energies. (Too lazy...)

(1) The potassium iodide has to be dissolved into the solution since you added it as a solid. This means there is an energy change and this could be endothermic or exothermic. Adding more salt increases the effect obviously.

(2) It takes some time to dissolve the potassium iodide. So before all the potassium iodide is dissolve some is already reacting with the hydrogen peroxide.

Greater surface area (smaller particles of potassium iodide) would allow the crystals to dissolve faster.

(3) The reactions are:
$\ce{H2O2(aq) + I-(aq) = H2O(l) + OI-(aq)}$
$\ce{H2O2(aq) + OI-(aq)= H2O(l) + O2(g) + I-(aq)}$

Overall the reaction is:
$\ce{2H2O2(aq) =  2H2O(l) + O2(g) + \text{heat}}$

But the heat is the combination of the two reactions. 

If $\ce{[H2O2](aq) &gt;&gt; [I-](aq)}$ then both reactions will happen.

If $\ce{[H2O2](aq) &lt;&lt; [I-](aq)}$ then only the first reaction will happen. The iodide isn't acting as a catalyst but as a reactant. 

3.3 % Hydrogen peroxide is 

$0.033 \cdot 15 \approx 0.5$ g $\ce{H2O2}$

Moles $\ce{H2O2}$ = $\dfrac{0.5}{34} = 0.015 \text{moles} = 15 \text{millimoles (mM)}$

0.2 g KI = $\dfrac{0.2}{166}$ = 1.2 mM

1.0 g KI = 6.0 mM

So you're adding so much KI that some might get left as $\ce{OI-}$. 

(4) so to do the experiment better:


cut the KI concentration
dissolve the KI in 5 ml of water &amp; equilibrate  temp to that of 3.3% H2O2
mix solutions with stirring