Since you're new to EE, I'm going to have to say that the answer is no. What you are asking is pretty straightforward, but it's not something you can do with hardware-store parts.
Basically, what you need is an energy storage device that you can insert between your car battery and your computer. This is complicated by the fact that you need to isolate this storage from variations in the battery voltage. That is, when you try to start the car and battery voltage drops, you don't want your storage device to try to backfeed the battery, as this will exhaust the storage device and your computer will die. In principle, you can do this one of two ways:
simulate this circuit – Schematic created using CircuitLab
Note that I'm showing your storage device as BOTH a capacitor and a battery. This is not necessary - you can use either. Using a battery instead of a capacitor is in some ways easier, but you have to worry about keeping it charged, which can be a royal pain in the whatzit. On the other hand, you don't have to worry about making sure the capacitor is charged before starting the car, which can be a problem, too.
Also note that the second approach, storing energy after the converter, seems simpler because you don't need a diode to keep from backfeeding the car battery, but this may or may not be true, depending on the converter. It assumes that the converter will not let current flow from output to input when the converter is not working. This may or may not be true. If you select a battery, it also assumes the converter will turn on properly with a humongous capacitor hanging off its output. This also may or may not be true.
Using a capacitor will work, but you have to size the capacitor properly. To do that you have to know a) the current it supplies when needed, b) how long it needs to provide the current, and c) what voltage drop is acceptable during that period. If the answers are a) i amps, b) t seconds, and c) DV volts, then
C = 1,000,000 x i x t / DV (in uF)
So, if you're supply 2 amps for 3 seconds, and you can accept a voltage drop of 2 volts,
C = 1,000,000 x 2 x 3 / 2
= 3,000,000 uf, or 3 F.
You'll have trouble finding a 3 F, 12 or 24 volt capacitor.
So, like I say, this may be rather more trouble than you can handle at this stage in your education.