Activation of multiple LEDs with one or several transistors
Both will work well. However, the second circuit is obviously more expensive to build (and takes more time to solder manually), so you will probably choose the first option (single transistor).
The second option (multiple transistors) can be a better choice when there is a lot of current going through the LEDs. You can then share the current through multiple transistors, which can therefore be smaller (sometimes, 4 small transistors are cheaper than one big, and the heat can be dissipated more easily). But given the currents involved in your case, it doesn't matter.
Either method works, but I would generally use the first because it is simpler and requires fewer parts.
If the signal is actively driven both directions, like from a CMOS digital output, then you don't need the resistor between base and ground. In other words, you can lose R2 in the first circuit, and R10, R11, R12, and R13 in the second.
As you say, the transistor in the first circuit needs to be able to handle the combined LED current. For four normal indicator LEDs, that is not much of a limitation.
Depending on the voltage of the LEDs, you might be able to drive two at a time in series with a single resistor. For example, if these are red with 1.8 V forward drop, and the transistor goes to 200 mV in saturation, then there is still 1.2 V left for a resistor to set the current. Doing that gets you the same LEDs lit with the same brightness, but with half the current used.
Green LEDs with 2.1 V drop are on the edge, but can work doubled up. Two LEDs would drop 4.2 V. Again, figuring 200 mV for the saturated transistor, that still leaves 600 mV for the current limiting resistor. Especially if you're not trying to run the LEDs at their limit, this can be a legitimate current savings.
Both circuits will work.
The first circuit is entirely adequate unless you value independent control of each LED.
The second circuit has the advantage that if desired you can turn individual LEDs on or off independently and the (obvious) disadvantage of requiring more components.
Almost any NPN transistor will work. If the LEDs are white they will have ABOUT 3v on voltage. So I per LED = V/R = (5-3)/560R = 3.6 mA per LED or about 15 mA for all 4 LEDs. Almost any small NPN transistor will work here.