I used 50mA and 100mA numbers off the top of my head, just to be able to give you some examples.
You could use just one bridge rectifier, or two bridge rectifiers, it's fully up to you.
Again, to make this easier to follow, let's say you need 48v at maximum 20 mA (you say 15mA but let's be safe and size it for 20mA) and let's say you need 24v at maximum 50 mA.
The way linear regulators work is by wasting away the difference in voltage in the form of heat. So, if you want a linear regulator to produce 24v at up to 50mA, that regulator will draw up to 50mA from the input, no matter what voltage is there.
So, if we want to use the 48v to get 24v from there using a linear regulator, it means that the 48v output must be able to supply both the 20mA and the 50mA, so in total at least 70mA. Just to be extra safe, we could round this number up a bit to.. let's say 75mA.
36v ac -> bridge rectifier -> capacitor - > 48-52v DC at up to 75mA ->
A bridge rectifier converts AC wave to DC wave, and you'll have a DC output but with lots of waves, voltage going up towards a peak equal to AC voltage x 1.414 (square root of 2) and then again down towards 0, up to 100-120 times a second, depending on your mains frequency.
In a bridge rectifier, there's always two diodes conducting so there is some voltage drop on each of those two diodes, but the exact amount varies with the current and how warm the bridge rectifier is. With regular diodes, at 75mA, the voltage drop on each of those diodes would be about 0.5-0.6v but if you were to use shottky diodes the voltage drop would be much low, at around 0.25-0.4v.
So, your 36v AC will become about 51v DC maximum, minus about 1v in losses in the bridge rectifier, less if you use shottky diodes. But let's make it easy and say we will have a peak of 50v DC.
But as I explained, 100-120 times a second the voltage will go down towards zero, and that's where that capacitor comes in... it stores energy inside it when the wave goes up to 50v and gives it back when the voltage goes down. So if this capacitor is big enough the voltage will never go down below a particular level.
Let's say we never want to have less than 49v. In this case, I already told you the formula C = 0.075 A / ( 2 x 50 Hz x (51v - 49v) ) = 0.075 / 200 = 0.000375 Farads or 375 uF which is the minimum we would need to make sure we'll always have at least 49v in a country with mains frequency of 50Hz (europe etc, with 60 Hz even less capacitance would be needed). So the next standardized value capacitor would work.
You can use a larger capacitor, for example 470uF or 1000uF, and that will simply make the voltage stay much closer to the peak of 50v instead of staying near the 49v value.
Now from here, if you really think you need 48v, you can use a linear regulator like LM317HV (which can handle up to 60v) to have the 48v.
As for 24v, you can use another LM317HV to get the 24v up to 50mA. The wasted power which is up to (48v-24v) x 0.05 = 1.2w is not that big to be worth messing about with another bridge rectifier and capacitor.