It is a basic rule. The physical basis for it is as follows. Current is just moving charges. Charges cannot accumulate. That is, objects generally cannot acquire a net charge. That is both a physical rule (rule of the universe) and also a rule for circuit analysis. So every time charges move into one side of a conductor, an equal number of charges has to move out on the other side.
So if I2 is dumping charge into the conductor at 10A, and I1 is clearing charge from the same conductor at 5A, that would imply that charge is accumulating in the conductor. Since that is impossible, it is impossible for two current sources to be in series unless they have the same current.
I hope you find this to be a satisfactory answer. I have tried to be accurate but intuitive. There are lots of analogies that could apply also (cars on a freeway, or skiers getting on a ski-lift or something). Let me know if you want me to offer something of that nature.
First, please don't try "supply fight" shenanigans in practice if you don't know what to expect.
Assuming both sources are exactly 10V, the current capacity wouldn't matter. Just use Kirchoff's circuit law as usual: 10V on one source, -10V on the other, so 0V on the resistive load. V=RI=0, I=0, no current.
Current capacity would matter if:
1) There was some voltage difference between the sources;
2) This voltage difference causes enough current to flow through the circuit so as to overload the current capacity of one or both sources.
In practice, there are probably no two sources that produce the exact same voltage, so condition 1 will always be true. However, condition 2 rarely will be satisfied: you need a large I, so you need a large V/R. If both sources are rated as 10V, they probably shouldn't differ much, so V will be small. And R will be at least the sum of the output impedances of the voltage sources.
If you actually want to explore how the current limitations would affect the outcome, you should consider what happens when 1 and 2 are true. I'd suggest assuming that one voltage source is +10V/10A and the other is, say, -9V/100A, and R is sufficiently tiny (say 0.05 ohms).
Best Answer
You seem to have some confusion regarding the voltage and current ratings regarding a voltage source. An ideal voltage source will have a constant voltage across it and will source or sink any amount of current to keep that voltage constant. In real life voltage sources cannot source infinite current, thus they have a current rating that represents the maximum current it can source before something bad happens. What will happen when you exceed the current capabilities of the voltage source depends on the device, it can blow a fuse, shut down, drop the output voltage or become damaged.
Regarding the circuit you have posted assuming point A is ground the voltage on the other side of V2 will be 5v, past V1 will be 25v and on the other side of V3 will be back down to 5v. Thus the voltage across R1 is 5V and that makes the current (via Ohm's law) 100ohm/5V = 50mA. If we go back and look at the current ratings of the voltage source, V1 and V3 are capable of 200mA and V2 can handle 50mA which means they can handle the 50mA current calculated above.
This is assuming that the current ratings on the voltage source you mention is for sourcing and sinking current, as in the case of V3 it is sinking 50mA of current, which may not actually work in real life depending on the specifications of the voltage source (many real life voltage sources don't like sinking current). If V3 is only capable of sourcing current and not of sinking current then the voltage source is being used outside of its specifications and all sorts of bad things may occur. It may have some sort of internal protection and thus not allow any current through, in which case there will be 25V across it. If it doesn't have any protection it may become damaged and act as a short and thus have 0V across it.
Also running V2 at the maximum current rating is ill advised as any small variation in the circuit can cause the current to exceed the rating of the voltage source.