The input capacitor reduces the impedance of the power feed as seen by the regulator. This reduces input voltage fluctuations that occur as a function of current demand fluctuations, which the regulator has no control over. The regulator can do a better job of keeping the output steady when the input is steady.
The regulator electronics are specifically designed to keep the output steady with the input varying over a wide range (that's the point of a linear regulator), but no such circuit is perfect. Note the input regulation spec in the datasheet. That tells you how much input variations are attenuated to the output. However, that is usually at DC. As the frequency of input variations goes up, the active circuit becomes less effective at attenuating them to the output.
Fortunately, a capacitor has lower impedance at higher frequencies, thereby reducing the voltage fluctuations due to current fluctuations at higher frequencies. The capacitor does a better job as the active circuit does a worse job. Another way to think about this is that the capacitor together with whatever impedance there is in the power feed to the regulator form a low pass filter. This filter reduces the high frequency voltage fluctuations that the active circuit is less good at dealing with.
Yes, you could add an entire extra linear voltage regulator and capacitors for a voltage rail you don't even need, just to spread out power dissipation.
Or you could just use a resistor.
You say your problem is reliability? Then let me tell you about the most reliable electrical component ever made: the resistor. They're the only component that can, albeit unreliably, continue to function even after catching fire. Heck, I've seen those beefy sandstone power resistors explode a little and still work, at least until they explode a lot.
But I digress.
Just put some resistors between the 12V buck-boost converter and your 7805. You can parallel several if you like to really spread out the dissipation and give yourself lots of head room. This will be the most reliable, simple, and inexpensive solution.
And, as a bonus, you will almost certainly get better ripple reduction than cascading an additional 780x (7809 in your case).
The 780x series regulators are fairly slow. They have ~60-70dBV of ripple rejection but only if that ripple is 120Hz. I don't know what frequency your dc/dc converter is operating at, but it is almost certainly 100kHz or more, and will have switching harmonics into the MHz. Those will shoot through a slow linear regulator like the 7809 like a bullet shoots through air.
Ok, it's not quite that bad, but I would be surprised if you got more than 40dBV of rejection at 100kHz, and if your buck-boost converter is faster, it could be 20dBV or even worse.
But do you know what you get if you have a resistor in series with a capacitor? A low pass filter! YAY! And if you placed a voltage dropping resistor before your LM7805, you very conveniently have just such a capacitor - the input capacitor for the linear regulator. Sure, the 7805 needs a nice low impedance input, but that's what the capacitor is for. And since you know your maximum load current, you its a trivial matter to size your voltage dropping resistor. Its really just a matter of how much heat you want to dissipate in the resistor(s) vs. in the linear regulator. The nice thing about, say, 1W resistors, is that they don't mind getting hot and will dump that heat with natural convection alone, so I would recommend favoring the resistors when it comes to shedding watts.
I'll just do a quick and dirty example. At 500mA load current, let's drop the voltage to exactly what you'd get with a 7809. To do that, we need a 6Ω resistor. If you want to be really safe, perhaps use 3 18Ω 1W resistors in parallel, then each one will only have to dissipate 500mW. They will drop the voltage to 9V before it even reaches the 7805. If you put a nice fat ceramic directly on the input of the 7805, perhaps a 47µF one or even a fancy 100µF one, in all its 0805 sized glory, you'll solve your problem in a very reliable way, AND have a beefy low pass filter that will give you better ripple rejection than an extra linear regulator ever could, especially at higher frequencies.
And if you really want to clean up the voltage even more, toss in a ferrite bead in series between the resistors and the input capacitor. Ferrite beads are wonderful little critters. Unlike inductors, which might reduce ripple but also radiate some of that energy out as EMI and almost certainly make the situation worse, ferrite beads take high frequency ripple and dissipate it as heat due to core losses. They're best thought of as frequency dependent resistors that only have resistance above certain frequencies. They will aid you a great deal when it comes to cleaning up the analog sections of circuits - now is the best time to start using them!
If you still really have your heart set on adding a 7809, then you can still solve your problem this way. Add a smaller series resistor between the 7809's output and the 7805's input. This will dissipate a bit of power and spread things out further, and will decouple the 7809's output capacitor from the 7805's input capacitor. Just use the values you would if the regulators were being used by themselves. The input capacitor is what provides the low impedance power to the voltage regulator, so it is perfectly fine to have some resistance in series with the input, as long as it is before the input capacitor. Never put it between the input capacitor and the regulators actual input, obviously.
Best Answer
You say "When the device is about to transmit the signal" and use the word "sudden" which implies things are going fast, so we can rule overheating.
Check 7805 regulator input and output voltages with a scope while doing your transmission test.
Some modern fast LDOs are happy with 1µF ceramic at the output. But 7805 is quite an old design, and as such it is slow to respond to fast current changes. Thus it needs capacitance on the output to smooth things out. Maybe your DC-DC is slow to respond also, but this is less likely (do the above measurement).
Since a 100µF 25V capacitor costs 5c you might as well put one on both sides.
You'll want to keep ESR under 2 ohms to reduce voltage sag due to 100mA current spike to a reasonable 0.2V maximum. Since most general purpose caps have tandelta around 0.1-0.2, this means 100-200µF or more. A 10µF cap would have too much ESR to be of any use on the output.
You could use a low-ESR cap if you have'em, but then you'd have to worry about stability and such, so basically just reach into your parts bin and grab any general purpose electrolytic that's big enough and fits. It won't hurt if it has more capacitance than required.
Also if the thing will sit outside in the cold, remember that capacitor ESR goes through the roof at low temperatures, so in this case you might want to oversize the cap or use something a little more evolved like Panasonic FC. Or just stick a 10µF ceramic in parallel instead of your puny 330nF.