For straight-up development boards (for your internal use), I go with a solder jumper or put two back-to-back (3 pads) to make an SPDT switch (here's a footprint I use). If it's small enough, it's fast to both close and open with a touch of solder or desolder braid. Using an actual resistor makes it much more difficult to rework with a standard iron.
If this is a product (as in, the Atmel STK500 development board is a product), you should use something like jumpers or DIP switches, because you don't want some dumb user poking around your board with a 1000°F iron. I'd tend towards DIP switches if you have more options or you are going to put it in an enclosure, otherwise jumpers would be cheaper.
The main question should be "is this something that will be changed as part of normal use?" If the answer is yes, requiring a soldering iron and skills is inappropriate. If it's something that an end user might modify 1-5 times (or preferably someone skilled, e.g. a lab tech), a solder jumper might be OK.
The impedance (think of it as resistance) of a capacitor changes with the frequency of the signal passing through. The lower the frequency (bass sounds) the higher the impedance.
The impedance of the capacitor also depends on its value. A capacitor with a higher value will have a lower impedance than a capacitor with a lower value. For the same frequency, a small valued capacitor represents more resistance than the large value capacitor.
In order to get more bass, you have to use a larger capacitor in series with the speaker.
C1 in your circuit is there to block DC from the amplifier. At DC, a capacitor is very close to an open circuit - DC cannot pass.
The change over is gradual, however. The capacitor doesn't just block DC. It also impedes the flow of other frequencies. The lower the frequency, the more it is blocked.
At some point it is no longer noticeable. For working with filters (the capacitor/speaker combination is a high pass filter,) this point is defined as the point where the amplitude is reduced by half (that's -3dB.)
I'm not going to get into calculating the cutoff of a filter - there's plenty of explanations on the web that go into much more detail than I want to.
For the other side (resistor changes sound,) we have to look at inductors.
The pickups on your guitar are inductors - basically just coils of wire.
Inductors are the opposite of capacitors. Inductors let DC pass just fine, but their impedance goes up the higher the frequency. It also goes up as the value of the inductor increases.
You aren't changing the impedance of the inductor (pickup.)
When you change the resistor at the amplifier, you are changing the load on the inductor.
A resistor that is connected across the inductor forms a voltage divider. How the voltage is divided between the pickup and the resistor depends on the frequency of the signal - the impedance of the inductor changes with frequency which changes how the voltage is split between the inductor and resistor.
The combination of the coil and the resistor forms a low pass filter. It removes high frequencies.
The point (frequency) where this begins to be noticeable depends on the resistor loading the coil. A higher value resistor allows more high frequencies to pass. Lowering the value of the resistor lowers the frequency at which you can hear a difference.
Another thing that will happen is that the resistor also changes the amplitude of the signal presented to the amplifier. A higher resistor means less signal getting to the amplifier, which results in a quieter output.
A lower resistance means more signal to the amplifier, which gives a louder output.
For a guitar player, there is also the interesting possibility of distortion. You provide so much of an input signal that producing the amplified signal would require more voltage than the power supply of the amplifier.
When that happens, the output voltage will "stick" to the powersupply voltage until the input signal is smaller.
This is known as clipping, and is a bad thing in a general amplifier, but can be a useful thing for a guitar player.
Best Answer
From a microwave engineers point of view, things really start to become a problem when the physical dimensions of the component are comparable to the wavelength of the signal you are considering.
The length of the pins onto which this jumper mounts are 0.230" (5.84mm) and the separation is 0.100" (2.54mm) so the signal path length is approximately 0.560" (14.22mm)
This dimension represents 1/10 of a wavelength at about 2.5GHz. To reasonably represent a square wave, I would include upto the fifth harmonic - so I would place an upper limit of 500MHz on this part. This is a heavily simplified first off starting point based on rough dimensions and rules of thumb and obviously you should verify the performance by either a full 3D EM simulation if you have the facility, or by direct measurement of a test piece using a network analyser.