Linear regulators don't require a lot of capacitance on the input.
You do need something to smooth out the ripple of the power adapter, and also because the long cable of the adapter can act as inductor, when the circuit's power consumption varies by some significant amount.
A formula approximates how much capacitance you would need when using classic transformers, not wallwart/switching power supplies.
Capacitance (Farads) = Current / [ 2 x Mains Frequency x ( Vdc peak - Vdc minimum desired) ]
In your case, you're producing 12v using a 7812 linear regulator. Such a regulator wants an input voltage 2v above the output voltage, so you want at least 14v.
Using the above formula, you'd use C = 0.1 A / 2x60x(15 - 14) = 0.1 / 120 = 0.000833 Farads or 833uF
so technically even with a classic transformer, you'd only need around 833 uF to always have at least 14v.
As you have a switching power supply, you'd probably be fine with a much lower value, like 220uF..470uF, whatever standard value you can find.
On the output of linear regulators, it depends on the regulator. Some linear regulators don't require capacitors.
Some say in datasheet that some amount is recommended, and some linear regulators also require that capacitor to have certain characteristics. For example, some 1117 linear regulators require a capacitor with ESR between 0.1 ohm and 1 ohm, which means those linear regulators will not work well with ceramic or polymer capacitors, as these have too low ESR.
In the case of these 7812 , it doesn't really need capacitance on the output, but it won't hurt to have something on the output. An electrolytic with value between let's say 10uF and 100uF will be enough.
If the digital part is not that picky, you could have saved a regulator by getting 5v from 6v output. A simple 1n400x diode would cause a voltage drop of around 0.7v..0.8v, so you end up with around 5.2v, and anything that works with 5v should tolerate 5.2v just fine.