This is pretty straightforward: your transformer only supplies current during a small proportion of the line frequency cycle. This is the period when the transformer output is greater than the capacitor voltage. When the transformer voltage is less than the cap, the bridge diodes prevent current flow. The transformer is unable to provide the relatively high-current, low-duration pulses required to keep the cap charged. Essentially, a bridge only provides current to the cap for a fairly narrow window around the peak of the input sine wave. If it's not, this implies that the output current from the cap is so high that the voltage is being kept much lower than you think.

Another issue is the size of your filter cap - it is far too low. You need to learn to calculate the voltage/current relationship of capacitors. The base equation is $${dV}/{dt} = i/C $$ so, for instance, if you are drawing 10 amps from a 2000 uF cap, when the transformer voltage is less than the cap, the voltage at the cap will change at a rate of 10/.002, or about 5000 volts/second. Assuming you are using 50 Hz line frequency, the cap will completely discharge from 15 volts in 15/5000 seconds, or about 3 msec. Since your line frequency is 50 Hz and the output of the bridge will be full-wave rectified sine at 100 Hz, the load will only have current through it for about 30% of the time, and the average voltage will drop significantly, since it will be zero about 70% of the time. To make things worse, the transformer will be unable to fully charge the cap during each cycle,so the average will be even worse.

All of this will be obvious if you have an oscilloscope. I suggest you get a cheap PC-connected scope, which have a fairly simple A/D connector board and use the PC as a display. These are available from dBay quite cheaply, although as always caveat emptor.

A simpler approach is to ditch your load box entirely. To see what the transformer can do, load it directly with some power resistors, and use the meter to measure both the load voltage and current (although not both at the same time, of course). Simply put a known resistor/resistors on the transformer and measure the voltage. Dividing the voltage by the known resistance will let you back out the current. Just be careful about cooling the resistors. A 15-ohm resistor on a 12-volt transformer will dissipate nearly 100 watts.

This is pretty straightforward: your transformer only supplies current during a small proportion of the line frequency cycle. This is the period when the transformer output is greater than the capacitor voltage. When the transformer voltage is less than the cap, the bridge diodes prevent current flow. The transformer is unable to provide the relatively high-current, low-duration pulses required to keep the cap charged. Essentially, a bridge only provides current to the cap for a fairly narrow window around the peak of the input sine wave. If it's not, this implies that the output current from the cap is so high that the voltage is being kept much lower than you think.

Another issue is the size of your filter cap - it is far too low. You need to learn to calculate the voltage/current relationship of capacitors. The base equation is $${dV}/{dt} = i/C $$ so, for instance, if you are drawing 10 amps from a 2000 uF cap, when the transformer voltage is less than the cap, the voltage at the cap will change at a rate of 10/.002, or about 5000 volts/second. Assuming you are using 50 Hz line frequency, the cap will completely discharge from 15 volts in 15/5000 seconds, or about 3 msec. Since your line frequency is 50 Hz and the output of the bridge will be full-wave rectified sine at 100 Hz, the load will only have current through it for about 30% of the time, and the average voltage will drop significantly, since it will be zero about 70% of the time. To make things worse, the transformer will be unable to fully charge the cap during each cycle,so the average will be even worse.

All of this will be obvious if you have an oscilloscope. I suggest you get a cheap PC-connected scope, which have a fairly simple A/D connector board and use the PC as a display. These are available from dBay quite cheaply, although as always caveat emptor.

A simpler approach is to ditch your load box entirely. To see what the transformer can do, load it directly with some power resistors, and use the meter to measure both the load voltage and current (although not both at the same time, of course). Simply put a known resistor/resistors on the transformer and measure the voltage. Dividing the voltage by the known resistance will let you back out the current. Just be careful about cooling the resistors. A 15-ohm resistor on a 12-volt transformer will dissipate nearly 10 watts.