You have the right idea for a basic unregulated supply. A transformer, four diodes, and as large a cap as you can manage will serve well enough for a lot of purposes, but isn't appropriate for all.
There are two main problems with such a unregulated supply. First, the voltage is not known well. Even with ideal components, so that the AC coming out of the transformer is a fixed fraction of the AC going in, you still have variations in that AC input. Wall power can vary by around 10%, and that's without considering unusual situations like brownouts. Then you have the impedance of the transformer. As you draw current, the output voltage of the transformer will drop.
Second, there will be ripple, possibly quite significant ripple. That cap is charged twice per line cycle, or every 8.3 ms. In between the line peaks, the cap is supplying the output current. This decreases the voltage on the cap. The only way to decrease this ripple in this type of design is to use a bigger cap or draw less current.
And don't even think about power factor. The power factor a full wave bridge presents to the AC line is "not nice". The transformer will smooth that out a little, but you will still have a crappy power factor regardless of what the load does. Fortunately, power factor is of little concern for something like a bench supply. Your refrigerator probably treats the power line worse than your bench supply ever will. Don't worry about it.
Some things you can't do with this supply is run a anything that has a tight voltage tolerance. For example, many digital devices will want 5.0 V or 3.3 V ± 10%. You're supply won't be able to do that. What you should probably do is aim for 7.5 V lowest possible output under load, with the lowest valid line voltage in, and at the bottom of the ripples. If you can guarantee that, you can use a 7805 regulator to make a nice and clean 5 V suitable for digital circuits.
Note that after you account for all the reasons the supply voltage might drop, that the nominal output voltage may well be several volts higher. If so, keep the dissipation of the regulator in mind. For example, if the nominal supply output is 9 V, then the regulator will drop 4 V. That 4 V times the current is the power that will heat the regulator. For example, if this is powering a digital circuit that draws 200 mA, then the dissipation in the regulator will be 4V x 200mA = 800mW. That's will get a 7805 in free air quite hot, but it will probably still be OK. Fortunately, 7805 regulators contain a thermal shutdown circuit, so they will just shut off the output for a while instead of allowing themselves to get cooked.
The first approach is to connect the 7809 directly to the positive output of the full wave bridge. The peak voltage there should be about 17 V. That means the 7809 would drop 8 V, which will dissipate 800 mW at 100 mA out. A TO-220 standing up in free air at room temperature should be able to handle that. Check the datasheet.
One nice thing about the 78xx regulators is that overheating won't damage them, they'll just shut down, cool off for a little while, then come back on again. You can therefore just try it and see what happens. Although I think it's unlikely it will get so hot as to shut down at only 800 mW, you can add a small heat sink if you run into problems. There are heat sinks specifically made for clamping or bolting to TO-220 packages.
The other possibility is to add the 7809 after the 7812. This will spread the dissipation across both devices. However, it decreases the current available from the +12 supply relative to the -12 supply, and could now cause the 7812 to overheat. With 100 mA at 12 V, and another 100 mA for the 9 V regulator, the 7812 will now dissipate up to (17 V - 12 V)(200 mA) = 1 W. Since that gives you a worse worst case than the other way, and makes the 12 V supplies assymetric, I'd put the 7809 directly on the unregulated 17 V line.
Either way the same overall power will be dissipated as heat at the same current.
Best Answer
Update: I checked that link and it seems you want to use a 3 VA transformer to make a 60 VA supply !? You'll need a transformer of 75-80 VA.
First of all, a voltage regulator at 5A produces a lot of unnecessary heat.
Sticking to your design, I'll assume the voltage regulator has a drop of maximum 2V accross it. So it must be powered all the time with a minimum voltage of 14V.
So you have that transformer of 12 V RMS AC. After bridge rectifier you get \$12 \cdot \sqrt{2} - 2 \cdot 0.7 = 15.57V_{p-p}\$. This voltage should never drop below 14V. This means a maximum ripple of 15.57-14 = 1.57V. For even better reliability I'll approximate this to 1.5V.
Now the capacitor value is \$C = {I \over {2fV_r}} = {5 \over {2 \cdot 50 \cdot 1.5}} = 0.033 F = 33000 uF\$.
This high value makes your transformer unusable for this design. Now that you know how to calculate this, choose a transformer with higher output voltage so that the filter capacitor will have a smaller value.