Building a power supply, Mysterious voltage jump?
My main question is why does the voltage jump so high after the
  capacitor and how can I fix it?


In simple terms, the capacitor acts as a peak voltage hold component thus charging to the top of the peak of the rectified secondary voltage. 



Without the capacitor, the average voltage measured from the rectifier (using a standard DC multimeter) would be about 0.637 x 30 volts x \$\sqrt2\$ =  27 volts.

The 0.637 factor comes from here: -



Here's the proof: -





With the capacitor present, the DC becomes "smoothed" at the peak value from the bridge rectifier which is around 42 volts (30 volts x \$\sqrt2\$). This is where the 0.707 value arises in the top picture i.e. \$\sqrt2\$ = 1.4142 = \$\dfrac{1}{0.7071}\$.



To get a more accurate result we should remove about 1.2 to 1.4 volts from the numbers above to account for the volt drop caused by the diodes in the bridge rectifier.

To restore the smoothed and rectified voltage to something like 26 volts DC requires a transformer change with a secondary voltage that is more like 18 volts RMS.

In other words, 18 volts RMS x \$\sqrt2\$ = 24.5 volts minus a couple of diode drops to result in a final voltage of about 23.2 volts. If you can find a secondary voltage rating of 20 volts then it would be more suitable.

Alternatively use a regulator but be prepared for issues such as excessive warming if the current draw is too much.