The "speed", rupture capability, and voltage spec of a fuse are each separate traits that aren't necessarily related. They will be specified according to the nature of the application.
Many cheap multimeters (i.e. <50$us) will only have glass fuses, because they're cheaper. But 'high rupture' capability (the ability to break a circuit up to a rated voltage, when a high fault current is flowing) is important in a multimeter (and in a circuit breaker), especially in a DC scenario, because DC >40V can easily form an arc and continue the current flow, even though the fuse has blown.
For the "20A" fuse in your meter, a HRC type would be wise, and rated for the max V rating of your multimeter, and same speed as the original.
The breaking capacity of a fuse specifies the maximum current which the fuse is guaranteed to be able to interrupt if a fault should occur.
The fuse you use must have a breaking capacity greater than the maximum possible current which can be delivered from the source it is connected to.
This has little or no relation to the amount of current normally consumed by the device being fused!
You haven't specified whether it is the input or the output of your buck converter which you are fusing, and the considerations are different for each case.
So if you're fusing the output, then the breaking capacity of the fuse should be greater than the maximum amount of current which your buck converter can supply.
If you're fusing the input to the converter, then the breaking capacity should be greater than the maximum amount of current which the source of that supply can deliver.
If your supply is a normal mains power outlet, then you can safely assume that there is a circuit breaker somewhere upstream of you and it is probably rated for something in the region of 20A - so your fuse must have a breaking capacity of no less than this.
If you use a fuse with a lower breaking capacity value than the amount of current which the source can supply, you risk the possibility that when an over-current condition occurs, you fuse will 'blow' but the current will arc over the blown fuse and continue to flow until it is manually shut off.
I discovered all of this the hard way much earlier in my career when I used a tiny fuse (100mA rating with a breaking capacity of maybe 50A) because the device I had built only drew a small current from the mains supply.
Unfortunately for me I didn't take into account that my tiny circuit would be connected to some very heavy industrial supplies and the one time something went wrong, the breaking capacity of the fuse I had chosen was woefully inadequate to interrupt the huge fault current (many 100's or even 1000's of A) - my little circuit was turned into a charred lump and the entire factory's electricity was shut off ...
Best Answer
To elaborate a bit on the answer by Neil_UK...
At a modest overload, the fuse wire will melt at its weakest point, and break the current.
At a larger overload, an arc will form across the ends of the broken wire. This arc will persist until more wire has melted and the gap is too long to sustain the arc.
At a massive overload, the wire will vaporize. The metal vapor will support an arc running the entire length of the fuse. This arc will persist until either something else breaks the current, or the fuse goes bang.
High current fuses are often sand-filled to help quench the arc, and have hard ceramic bodies, rather than glass, to resist the explosion.
Addendum, after some comments on the question and the other answer.
Ideally, the fuse should be rated to break the maximum prospective fault current for the circuit it's protecting. That is, the maximum current that could flow if you put a dead short across the output of the fuse, taking into account the size of the supply transformer and all of the cabling back to that transformer.
Sometimes that isn't practical, and you have to rely on upstream fuses blowing in the most extreme short-circuit cases. That can be acceptable if you know that the upstream fuse will blow before the downstream one fails catastrophically.