Simple answer: You would get 5V in your case, but referenced to -12V instead of ground. In other words, you would have -7V, not -5V. In addition, the regulator would only source current onto the -7V rail, not sink it as would be expected for a negative voltage.
If you want to run some circuitry from 5V between -7V and -12V (the -12V will be the ground for this circuitry), then you can use a positive regulator as you described. If however you want to run some circuitry between ground or higher and -5V, then you need the negative regulator.
Using that series resistor would be a bad way to regulate voltage, but I assume you do this to keep the 7805 from dissipating too much. But 400 mA \$\times\$ 60 Ω is 24 V drop across that power resistor, so at 24 V input you won't have nothing left. Decrease the resistor value to maximum 35 Ω, then 400 mA will cause a 14 V drop, and you'll have enough left to feed the 7805, which needs 8 V in. If you expect more current to be drawn by your circuit you can calculate the resistor value as
\$ R_{MAX} = \dfrac{24 V - 8 V}{I_{MAX}} \$
But for the 4050 you use a voltage divider to get 3.3 V. Don't. The voltage will vary widely with the load, and you want you power supply to behave predictably, i.e. be rock steady. Use an LDO regulator to get your 3.3 V from the 5 V, like the MCP1700, which is cheap, and has a low ground current.
edit after you uploaded the schematic
A number of things are not quite clear to me. I couldn't find the schematic of the Adafruit breakout board, but it seems to be 5 V, then why do you need the HEF4050 buffer, and why do you use a 3.3 V supply for that, if both breakout board and AVR are 5 V? If the breakout board would run at 3.3 V, like when it has an LDO on board, then I would suggest to run the AVR on 3.3 V as well, so you won't have connection problems. If you need the 7805 for the breakout board you can use the MCP1700 I mentioned earlier to power the AVR from.
R1 and R2 don't seem to serve a function. You're not using the pin as reset pin (make sure you enable the internal reset), then program it as output, and you won't need a pull-up. (I don't see why you use two 22 kΩs here instead of one 47 kΩ either.)
Best Answer
The output capacitors are because the 7805 can only supply a limited peak output current (average or transient) whereas a capacitor is only limited by its internal resistance. For instance, an electrolytic capacitor might have an effective series resistance of 1 ohm and when connected across a 5V supply, clearly the peak current it can supply will be about 5A - this is much bigger than what the 7805 can supply BUT remember we are talking transient demands not average demands by the load.
As a footnote, it's always best to put a capacitor like this where it is needed - at the point of load because transients of several amps up and down long copper traces can cause other problems.
The circuit you posted doesn't show a particularly important capacitor this being the input cap to the 7805: -
(source: antihero.org)
Note the 0.33uF cap shown at the input - if you have long feed-wires connecting the power to the 7805, a capacitor helps peak demands taken by the 7805 - this is because feed-wire has inductance and the capacitor prevents instabilities.
The LT1528 is typical of some "adjustable" voltage regulators - the output can be set by a potential divider from output pin to the feedback/sense pin. Normally the LT1528 runs at 3.3V on the output and the sense pin can help the 3.3V be reproduced at some distance from the regulator (say close to a load). If you look at page 1 of the datasheet it shows you various settings for these resistors that allow an output greater than 3.3V.