Let's explain how the filter capacitor works.
I have suggested 1,000u for a 1 amp supply.
A 1,000u will perform like this: when you are drawing 1,000mA, the ripple will be 5v.
That means the minimum voltage across the electro must be 8v as the 7805 requires 3v across it to operate the electronics inside the regulator.
8v + 5v = 13v.
The transformer must deliver 13v + 2v across the rectifier = 15v.
A 12v transformer is 12v AC and this will produce 12v x 1.4 = 16.8v
This will be suitable for the project.
Now, the losses in the 7805 when 1 amp flows, will be 14.8 - 5 = 10 watts
What Klaus is adding is this:
When the input voltage is much higher than needed by the 7805, the 5v ripple is above the voltage needed by the regulator and we don't have to worry about it.