How is a BJT a current source?

We might use the term "current source" or the term "current sink" depending on which way collector current flows. We might use one term for PNP BJT's and the other for NPN BJT's. But the emphasis of OP's question is not on direction, but more - why emphasize current rather than voltage for collector.

The example given of a BJT having a current gain of 10, with 2mA applied to the base results in the collector trying to sink 20mA from collector to emitter - which is assumed at 0V (gnd). The 2mA base current also adds to emitter current. But let's concentrate on collector current and voltage.

^{simulate this circuit – Schematic created using CircuitLab}

A 1k collector resistor connected to a 5V voltage source supplies collector current to the BJT. The 0.02A collector current that tries to flow would cause a voltage drop across the 1k resistor of 20V...that's not going to happen since the collector cannot go much below base voltage. The best that the BJT collector can do is to pull that 1k resistor down to approach 0V. Nearly 5mA could flow from collector to emitter - no more. Total emitter current would be 5mA due to collector plus 2mA due to base. In this scenario, I'd hesitate to think of that BJT's collector as a current sink. That collector is pulling down hard toward ground and the 1k resistor is too feeble to resist.

Now let's replace the 1k collector resistor with 100 ohms. This resistor will more strongly pull toward the +5V voltage supply. If you connected this 100 ohm resistor to GND, 50mA would flow. But the BJT can only sink 20mA, limited by its current gain. The collector will sit at 3V. At this bias point, I'm happy thinking of the collector as a current sink... If I replace the 100 ohm resistor with 200 ohms, 20mA collector current can still flow, and collector voltage will drop from 3V to 1V.
To the extreme, a collector resistor of zero ohms would result in collector voltage equal to the +5V supply, but 20mA collector current would be flowing. This behaviour is that of a current source (or sink). The OP's view of the transistor acting as a current limiter is valid only for the scenario where collector resistor is below 250 ohms (approximately).

The description of "current source" or "current sink" describes what happens when the transistor is in active mode fairly accurately.

^{simulate this circuit – Schematic created using CircuitLab}

As you can see for Vce from a few hundred mV up to 40V the collector current is fairly constant.

If your issue is that the current sink does not work without some Vce to put the transistor in the appropriate region of operation, that is correct, but then we'd have to consider a voltage regulator not a voltage source but a "voltage limiter" since it requires some input voltage and below the minimum it will provide less output voltage. Amplifiers, as well, typically redirect energy from their power supply to the output.

A BJT is not a current source. However, there are some circuits where a properly biased BJT can substitute for a current source. That is, a current source and a properly biased BJT will behave similarly in certain constrained contexts. They will NOT behave similarly in all contexts, because a current source is a source of power, while a BJT is not.

For example, in the schematics of operational amplifiers, one might see something like this

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The operational amplifier would not actually contain a current source, but would instead contain a properly biased transistor, like so.

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where \$V_{bias}\$ is set by the voltage across a diode connected transistor with a known (or approximately known) current. That is, Q3 is an output driver in what is perhaps a multi-output current mirror.

So why can we use a transistor here in place of the current source in the original schematic? Because the function of both is to cause a some amount of current to flow from the emitters of Q1 and Q2 to ground. Since that is the only thing we expect from whatever component is placed there, and since both a current source and a BJT will give us that desired behavior, we can substitute a BJT for the current source in that circuit.

For a more detailed understanding of why the BJT will cause a certain amount of current to flow from the emitters of Q1 and Q2 to ground, it may help to consider the operation of a current mirror. Q3 is almost certainly biased by another transistor (or possibly a more complex circuit) and together Q3 and that other transistor or transistors form a current mirror.

Here is the simplest form of current mirror.

^{simulate this circuit}

Some known or approximately known current is fed through Q1, and so a voltage is developed at the collector of Q1. That voltage is used to bias the base of Q2. If we assume that the current into the base of Q2 is negligible, and we assume that Q1 and Q2 are well matched, and we assume that the circuit is such that there is a voltage source of some kind that can supply current to the collector of Q2, then we can assume the current through Q2 has the same magnitude as the current through Q1. Because our assumptions are actually only approximations, and because transistors do not have perfectly flat \$I_C\$ vs \$V_{CE}\$ characteristics for a given \$V_{BE}\$, the current through the collector of Q2 will only approximately equal the current through the collector of Q1. But for many purposes, it is "good enough". Various additions to this simple current mirror can make it more accurate.

In summary, a BJT is not a current source, but in many circuits, a properly biased BJT can substitute for current source. Whether that substitution works depends on the circuit where the substitution is made. In some cases the substitution won't work. In many integrated circuits it does work.