Looking at a diagram, is often a good way to understand a problem. Below is a simple diagram that shows the fault current path.
You can see that the fault current will flow through all the breakers, and return to the source (the transformer) along the grounding conductor.
You might be inclined to think that the fault current will be extremely high (tens of thousands of amperes). However, due to the fact that wires have resistance, the current might be surprisingly low.
Without knowing the exact length and size of each wire, it's not possible to approximate the resistance. If you did have that information, you could calculate the resistance. With that, the voltage, and Ohm's law, you could calculate the fault current.
For this example, we'll assume the fault current is less than 100 amperes.
Because the current is not above the instantaneous trip level of any of the breakers, the short-circuit protection of the breakers will not trip. However, circuit breakers also have thermal protection, which opens the circuit based on overheating caused by current flow (overcurrent).
Each breaker will trip according to it's trip curve, based on multiples of current over the rated current of the breaker. Basically, as the fault current flows, the thermal protection device in each breaker begins to heat up (along with all the wiring in the circuit). Obviously the larger breakers (100 & 200 in the diagram) can handle more heat (current), so they're going to be able to handle the less than 100 ampere fault current.
Since the fault current is higher than the rated current of the smaller breaker (20 in the diagram), the thermal protection device will begin to overheat. With such a high current flowing through it, the device will likely open within a few seconds. However, if the current was lower, it could take much longer to trip (even minutes).
If we take another example, where we've figured the fault current at 150 amperes. Even though the current is now higher than the rated current on three of the four breakers, the smaller breaker is likely still going to trip first. This is because the time before the thermal protection of a breaker trips, is based on the the amount of fault current above the rated current.
The fault current is only 1.5 times higher than the larger breaker, but 7.5 times higher than the smaller breaker. Because of this, the smaller breaker will trip sooner. Thermal protection is designed this way, so that loads can draw over the rated current, but only for a limited amount of time. This allows things like motors to start, without tripping the breaker.
In most real world applications, the smaller breaker will trip first. If the resistance of the fault circuit was low, it's possible for the fault current to be above the instantaneous trip level of all the breakers. In that case, the first breaker (main) will likely trip first.
If any of the breakers are GFCI breakers, and the fault is to ground, the GFCI breaker will trip first.
More realistic example
This example will use the diagram above, but will attempt to estimate a more realistic fault current. We'll say that there's 100' of 3/0 CU. wire from the pole to the main service panel lugs. 3/0 CU. is 0.0000766 ohms/ft., so that's 0.00766 ohms.
0.0000766 * 100' = 0.0076 ohms
Next there's 50' of 3 AWG CU. from the main panel feeder breaker, to the second panel main lugs. #3 CU. is 0.000245 ohms/ft., so that's 0.01225 ohms.
0.000245 * 50' = 0.01225 ohms
Next there's 25' of 12 AWG CU. from the second panel breaker, out to the fault. #12 CU. is 0.00193 ohms/ft., for a total of 0.04825 ohms.
0.00193 * 25 = 0.4825 ohms
Now that we've reached the fault, the current has to follow back along the grounding conductor. The grounding conductor is made up of 25' of 12 AWG CU., 50' of 8 AWG CU., and 100' of 3/0 CU. back to the pole.
0.00193 * 25' = 0.04825 ohms
0.000764 * 50' = 0.0382 ohms
0.0000766 * 100' = 0.00766 ohms
Totaling up all the resistances, we end up with 0.16227 ohms.
0.00766 + 0.01225 + 0.04825 + 0.04825 + 0.0382 + 0.00766 = 0.16227
Using Ohm's Law, the fault current can easily be calculated using the formula current = voltage / resistance (I=E/R).
120 volts / 0.16227 ohms = 739.5 amperes
739.5 amperes of fault current.
That's 3.695 times the 200 ampere breaker, which according to a random trip curve I looked up, should trip the breaker between 8-25 seconds. It's 7.395 times the 100 ampere breakers, which would trip between 2-7 seconds. It's about 37 times the 20 ampere breaker, which is likely beyond the instantaneous trip current level.
In this example, the 20 ampere breaker will trip first (unless any of the other breakers are GFCI breakers).
Best Answer
Given your fairly precisely stated 15/16" OD of the pipe, the only pipe anywhere near that size is EMT "Electrical Metal Tubing". But you say it's plastic. I can't find any plastic pipe near that dimension. Everything is <7/8" or well over 1", nearly 1-1/16". I'll proceed assuming EMT, since my wire reco has a lot of slack.
It also sounds like you consider the 30A an undesired restriction, and would prefer to exceed it. That's just as well, because the cables now in the pipe are totally unworkable, and must be removed. I won't get into a codevio list, because that would be tiring, insult your father's work, and just cheese you off. Suffice it to say, there is no way I can possibly see to make use of the NM cables in the pipe.
You can run up to #6 copper, for 70 amps.
That 3/4" pipe will support up to 6 AWG wire - that is to say, three #6 THWN-2 individual wires, and one #8 bare ground wire. (even though EMT is a valid grounding path, I would not trust it outdoors). That will put you at 29% fill, which is perfectly reasonable for 4 wires in a pipe.
(this works for all pipes except 3/4" schedule 80 PVC; for that, a #8 ground wire won't fit due to conduit fill limitations; the #10 ground limits you to 60A. But then, if it was schedule 80, I supremely doubt you got two #10 cables in there!)
That means you'll be able to support the load you have planned, plus a pretty good sized EVSE, wood shop, whatever will come up for you.
Given the short run of the wire, I don't see any reason to bother with aluminum wire (#6 Al would limit you to 50A) or with shrinking the wire to save a few bucks. Having 70A service at the garage would be sweet.
The 10/2 NM cables have to go
Unfortunately they will be annoyingly difficult to pull. This is what you are up against.
2 x 10-2 inside EMT:
2 x 10-2 inside Sched 80 PVC: 
It's actually worse than it looks in the drawing. It'll be bind-up city. This is why oval cables are treated as a round wire of the large dimension.
But steady on; get that wire outta there, and the three #6 will go in happy as a lark. Or smaller wire will obviously be easier still.
As to your questions:
Feed breaker size
What breaker for the existing 10-2 cable? None. That cable can't be supported for a variety of reasons.
What breaker for new 3x#10 THHN wires? 30A. Only. Because of NEC 240.4(D).
What breaker for larger THHN wires? Since you are 100% in conduit, and using THWN-2 wires, look at the 75 degree C column in Table 310.15(B)(16) and round up to the next larger breaker. So 8Cu/6Al = 50A. 6Cu = 65->70A. Larger wires won't fit.
Subpanel size: No place to scrimp!
What size subpanel is appropriate for a 10 AWG feed? Subpanels are about breaker spaces, first and foremost. That is the defining characteristic of a panel. A lot of people decide they want to scrimp on the panel spaces, so they get a smaller panel. This saves them the price ... of a latté. Then later, they have a $1000 problem because the panel is full. So gospel around here is Think Big. Think at least 16-18 space, and a lot more if you reasonably expect to fill that. Spaces are cheap, in fact, they make other things cheaper. Look for combo packs which include a main breaker and some "bonus breakers" to save yourself some money.
The second characteristic is bus amperage / main breaker trip. "Going Big" probably mooted that, as we're way beyond the 30A-bussed or 60A-bussed micro-panels. So if your panel has a 100A main breaker, or a 125A bus, that'll work out just fine. The main breaker is only there to be a disconnect switch (since this is an outbuilding), it's not there to be a breaker, so we don't care what size it is.
Hurry up and pull that permit before NEC 2020 kicks; that'll require an exterior disconnect switch also. If you're in Massachusetts, too late.
"But I want the nearby breaker to be the same size so it'll trip locally instead of at the house!" That trick never works, and you shouldn't even be trying. First, both breakers see exactly the same amount of current, so it would be a 50/50 chance even if they were identical. Second, they're not identical - main breakers tend to have a more forgiving trip curve. Third, if that happens often enough to care, the feeder is too small and must be enlarged. (see why I'm recommending #6? :) Feeder must be at 25% above expected practical load.