The super quick answer is: None of the parts that you have selected are appropriate for switching line-level (headphone level) audio signals.
Off Topic Rant: It is often ill-advised to restrict the parts or techniques used in the answers. I covered this briefly in an answer on meta.EE.SE: Does EE.SE have a problem with the treatment of newbies? The old saying goes, "If all you have is a hammer, then everything looks like a nail." Currently, you only have a hammer. But you have no nail. Get the right parts and you will be much happier.
Long Answer:
The main issue that you have is that you want to switch a bipolar signal (a signal that has voltages that can be positive or negative), and you have limited power rails to use (+6v).
The Bipolar Junction Transistor, in this case the BF199, is not going to work. Ok, if you used enough of them, in a particular configuration, then maybe. But I wouldn't wish that on an EE with 20+ years of experience, and certainly wouldn't suggest that for a novice.
The MOSFET approach could be made to work (as Dave Tweed) suggests. But, there is a catch. Let's say that your audio signal can vary from +2 to -2 volts and Vgs(th) Max of your MOSFET is 4 volts. Then the gate voltage that you put on your MOSFETs must switch to +6 and -6v. The reason for this is when your switch is ON, you do not want the reverse body diode of the MOSFET to be conducting any current. And for that to happen, you need to have your MOSFET to stay on for any possible voltage of the audio signal.
If your gate voltage is less, the MOSFET might be turning on and off and causing the diode to conduct. Because the switching time of the diode is not zero, and the diodes are really crappy diodes, there will be some distortion added. The amount of distortion will depend on the MOSFETs used, and is really hard to estimate. The resulting audio could be "telephone quality", or might be reasonable for the average listener. In general, the smaller and faster the MOSFET the less distortion you will have. The two MOSFETs you selected are not small or fast.
So, you could get the MOSFETs to work, but you will need + and - power rails that are probably different than what you have available right now.
The other issue with your MOSFETs is that they are just huge. Physically. You will need four of them to switch one stereo signal. If you are muxing several channels together then you will need 8 or more. That's a lot of MOSFETs.
If we consider solutions that are outside of your selected MOSFETs or BJTs: Then an analog switch chip such as what Dave Tweed suggested, or similar ones by Maxim Semi, are good solutions. Pay attention to the On Resistance of these parts because that could be relatively high (30+ ohms for the cheaper ones). But otherwise, these chips are easy to use and effective. Relays are also good, especially when audio quality or a low on resistance is required. Latching relays could decrease power requirements by a lot. Another solution is to use a J-FET. J-FETs are the cheapest solution and have good to excellent audio quality, but are difficult to control because they require a huge voltage swing on their gates in order to turn on/off correctly.
If you can get away with a relay, I would go for that. Easy to use, super high audio quality, and mostly bullet proof. The down side is higher power consumption and not suitable for mobile applications (shock and vibration). My second choice for you would be an analog switch. Good audio quality and easy to use. A distant third choice are the J-FETs. Hard to work with, good audio quality, and inexpensive. MOSFETs are fourth. And BJTs are a super distant fifth choice.
http://www.digikey.com/product-search/en?lang=en&site=US&keywords=tip120
TIP120 is a BJT, which is a different family of transistors from FETs. Below is a broad, oversimplified, cartoon version of how they both work. The below assumes NPN and NMOS, as specified in the question. PNP and PMOS would invert some of this.
A BJT has very low base impedance; essentially, there's a diode between base and emitter. This means that if the transistor is "on", the base of the transistor will be ~.7V above the emitter. If you try to drive the base higher than that (say to 3.3V or 5V with a microcontroller I/O pin) an undesirably large amount of current will flow, and bad things will happen. You have to have something between the I/O pin and the transistor base to limit that current. Thus the resistor. The processor side of the resistor goes to 5V (or whatever your microcontroller logic rail is), and the transistor side goes to ~.7V. This voltage differential, divided by the resistance, gives you the current being injected into the base. That, plus the transistor characteristics, tells you how much current can now flow through the BJT collector-emitter.
A FET has very high gate impedance, so no current flows into the gate when it's turned on. You apply voltage between gate and source, and the "switch" closes. The gate can typically go up to 20V above the source, so driving a FET with a microcontroller isn't typically a problem.
Instead, you have the opposite concern: some FETs need more gate voltage than some processors can supply!
Now, there are all sorts of additional details. Sometimes you put a resistor in series with the gate of a FET, for filtering purposes. There is actually current flow into the gate of a FET, particularly at turn-on and turn-off, which can matter for some applications. And BJTs and FETs can be driven in an analog mode, where they're neither fully on or off, but somewhere in between. Sometimes that's good, sometimes it's bad.
When I'm wearing my microcontroller hat, I tend to use FETs wherever possible. In general, they're easier to work with and their losses are lower. BJTs are sometimes cheaper, and they're more likely to be the choice for analog control applications.
Best Answer
The post you link to does explain this, but in case it needs repeating and perhaps backing up with a textbook reference, the saturation region for a MOSFET is called so because the drain current saturates, i.e. basically stops increasing as Vds increases further.
You are correct that the active region of a BJT corresponds to the saturation region of a MOSFET when these devices are used as amplifiers.
The saturation region of a BJT (e.g. when turned on as a switch) corresponds to the triode/ohmic region of a MOSFET.
Some authors also call the saturation region of a MOSFET the "active mode", which does match the terminology used for BJTs. But they also call the triode/ohmic region the "linear mode" which perhaps doesn't help that much because "linear" suggests an amplifier rather than a switch. Linear here again refers to how the MOSFET characteristic looks like in that region rather than any external/use considerations. (Luckily, it seems nobody calls the BJT saturation region "linear mode".)
The only thing that's not confusing about this terminology is the cut-off region, which is the same for both. Here's a summary table for the correspondence (from an external/use viewpoint):
This summary also includes the reverse active region for BJTs, which is seldom used, but it doesn't include synonyms for the triode region; as I said "linear mode" or "ohmic region" are also used to denote the MOSFET triode region.