A titration mixture having pottasium oxalate and pottasium hydrogen oxalate!
You are right, $1$ mole of $\ce{KHC2O4}$ reacts with $1$ mole of $\ce{KOH}$. From that first piece of information in your exercise, you can find out that, in $\pu{10mL}$ of solution, you have $\pu{0.05 * 0.01 = 5 * 10^{-4}}$ moles of $\ce{KHC2O4}$. The next piece of information is meant to let you how much substance there is in the solution. $10$ moles of both substances react with $4$ moles of acidified $\ce{KMnO4}$. Which means the sum of the quantities of both substances in that solution is $\pu{(0.06 * 0.01 * 10) / 4 = 15 * 10^-4}$. So you got $\pu{a = 5 * 10^{-4}}$ and $\pu{a + b = 15 * 10^{-4}}$. Do the math.

Please tell me if I was clear enough.