Here is a simple topology. This is opimized for simplicity at the expense of efficiency and effective use of the transformer's capacity:
With just a bare transformer secondary, you only get half-wave rectification to make each of the + and - rails. A center tapped secondary would allow full-wave rectification of both rails.
For a short time once per power line cycle, D2 conducts and charges up C2. Likewise, D1 conducts on the opposite polarity of the AC voltage and charges up C1. C2 and C1 hold up the rails between the bursts of current delivered thru the diodes. The two regulators then produce the nice and clean +12 and -12 volts. C3 and C4 are small ceramic caps that work better at high frequencies than the larger electrolytic C2 and C1. These help keep the regulators stable. C5 and C6 are generally required by such regulators, and again help keep them stable.
The only real brain cycles required here is to calculate what C2 and C1 need to be. At 60 Hz, each will be charged up every 16.7 ms. For sake of analisys, let's simplify this to assume C2 is charged up instantly to the full voltage once every 16.7 ms with nothing being added to in at other times. That's actually a little worse than what really happens, so is a good thing to design to.
You say the maximum output current is 500 mA. A 15 VAC transformer will put out 21.2 V peak. Let's be pessimistic and say the diode drops 1 V at the full current. That leaves 20.2 V that C2 gets charged up to. 78xx regulators need a lot of headroom, let's say 2.5 V (your job to check the datasheet). That means the voltage on C2 can't be allowed to drop below 14.5 V. 20.2 V - 14.5 V = 5.7 V that C2 is allowed to drop in the 16.7 ms between getting recharged. (500 mA)(16.7 ms)/(5.7 V) = 1.5 mF. That should be at least a 30 V cap. The same logic holds for C1.
This should all work, but you also need to look at the power dissipation. This simple circuit won't be very efficient. From above, we can see that the average voltage on C2 at full current is 17.4 V, which means each regulator is dropping 5.4 V. That times the 500 mA current is 2.7 W each. A TO-220 package can handle that, but will require at least a small heat sink.
I don't understand what is happening on this diagram.
Referring to the diagram in OP.
This MOSFET-TRANSISTOR duo is used to enable and disable the power supply of the module, when MCU_CTRL is high Q101 is enable, which in turn enables the Q102 by pulling the GATE of the PMOS Q102 to GND, and enable the power for module on VBAT pin. When MCU_CTRL is low, in similar fashion it disable the power in VBAT pin.
For one thing, the part I was most worried about, the capacitor, seems not to be listed. The description of the diagram states that "A low ESR tantalum capacitor is usually used. The value for the capacitor should be more than 470uF."
The datasheet of SIM900 suggest low ESR 100uF capacitor(tantalum will be good) with a small 0.1uF or 1uF ceramic in parallel placed close to the SIM900, on page 20 of datasheet.
It also says that "When there was no on/off pin in the LDO or DC-DC IC, the customer can follow nether reference schematic to control the VBAT on/off." which seems to imply this doesn't actually show all of what goes into the VBAT and there's more to the circuit outside the VBAT_IN.
This particular section in the datasheet gives you an option to enable/disable the power supply of SIM900 using the MOSFET-TRANSISTOR duo if in case there is no ON/OFF pin present on the DC-DC converter or LDO. So either optios you can use to enable/disable the power supply of SIM900. But the datasheet of SIM900 shows it has got PWRKEY which is default pulled-up to enable or disable the power supply. Ref: Datasheet page 17.
I'm also confused as to what the MCU_CTRL is
The MCU_CTRL is external pin coming out from the master controller/processor, not coming from SIM900 to enable/disable the power supply.
In the starting of the OP, it disusses about powering the SIM900 using RBBB arduino which can supply a maximum of 250mA to 300mA which is using L4931 LDO.
But I am afraid it is not the correct approach to power SIM900, since the datasheet clearly expects to power SIM900 using LDO or DC-DC Converter which is capable of 2A current, with a minimum voltage of 3.4V. Ref: Datasheet page 20
Best Answer
LDO's have low output impedance but if the pulsed current exceeds the regulator rating, then additional storage capacitance is needed where C=I*dt/dV.
Generally high power audio systems do not use cheap simple LDO's but for low power , your approach is good. Bipolar supplies allow direct coupling down to DC if you want.
Generally I choose input storage cap and load R such that RC = 8/f for 10% max ripple voltage (Rule of Thumb) at input frequency , f.
Thus for you f=100Hz , if say load worst case is R=1k , \$C=\frac{8}{f*R}\$ = 80 uF (microfarad) and you have 1mF=1000uF so this is overkill. You could use 220uF for good results and anticipate <5% input ripple.
Beware that LDO's have a 2V drop depending on load current. (check specs) And bridge rectifiers raise no load voltage by 1.414 minus 2 diode drops and +10% allowance for transformer voltage with no load
There are other ways to calculate this using \$C=I * \frac{dt}{dV}\$ with \$I=\frac{V_{{in}_{min}}}{R}\$ so \$C = \frac{Vmax * dt}{R * dV}\$ so you choose you Vmax , dV for duration of a about 1/3 of a cycle of your lowest bass frequency without loss of voltage regulation.
Misc info
All dielectrics have this equivalent circuit which results in a self resonant frequency where Zc(f) is lowest. By shunting caps of smaller values extends this low impedance at higher frequency for the range required.
For e-caps or electrolytics which have highest density ( yet only 0.1% of a battery F/volume) but much higher ceramics they tend to have a much lower SRF but much higher C. A useful thing to rememmber is the quality of any electrolytic may be judged by its ripple current rating and better by it's ESR. For any family of caps and brand you will find in certain voltage range that ESR*C=T is constant. For the best e-caps used in SMPS, these are as low as 1us and the best microwave ceramics this will be <<1ns but for audio you don't need this bandwidth but you do want the drivers to have very low source impedance which is in series with your power regulators and input Caps. So for source impedance of 1% of load or an Audio woofer dampening factor of 100x on a 8 Ohm load you want Caps with an ESR < 80 mΩ and possibly lower so watch out for this value.
Now if the very best e-cap is T=1us and the cheapest GP cap is T=200us , which do you get? You need large supply high bass current with low ripple and you need small ceramic or plastic caps for high frequency pulses since they have more bandwidth, but since Zc(f) drops with rising f, you will often see 1000uF//10uF//1nF or variations of this depending on your power. For 100W you might see 100mF//100uF//1uF//0.01uF but it depends on the ESR and cost of each component and quality of the audio system design.
Now you will find General Purpose e-caps (Cheap) will have ESR*C = T = 200 us or in milliseconds in massive size and many seconds in batteries. We also know BW=0.35/T for rise time so a cheap cap has less effective Bandwidth.
You will find a lot of other resources on the net and in this forum about the unique characteristics of every dielectric material, that tradeoff density, ESR, temp stability, cost, size, voltage rating, polar or non-polar etc. Non-Polar e-caps are fundamentally two polar caps in series (+- -+) that may have reverse diode protection.