Ideally the LEDs should be driven by a constant-current circuit. This will maintain a constant brightness and color as the battery drains, or as the LEDs heat up or cool down.
But the real world isn't ideal, so you can often get away by just using resistors. Yes, you should definitely use them. The resistors are there to limit the current through the LED and keep them from overheating and burning up. A 9V battery has a fairly high internal resistance, so you may be able to get away with two in series and no resistor, but it will be unreliable (changing to a different brand of battery could be enough to blow out the LEDs, etc.)
For the worst case of two white LEDs in series running at 20mA, the lowest forward voltage shown in your link is 3.2V, so you would have (9 - 6.4)/.02 = 130 ohms. The current is low, so a 1/4 watt resistor will be fine. Select the closest value to this you can find. Running at 20mA the LEDs will be pretty bright and this is a benefit: as the battery drains or the LED forward voltage changes, the apparent brightness probably won't change that much. Human vision is more sensitive to dim lights and it's harder to tell that a bright light has changed 10% than a dim light has changed 10%.
The power supply is 3v at 60mAh (CC/CV)
A power supply can't force both the current and voltage to specific values at the same time. It can either force a certain voltage and let the load determine the current, or force a certain current and let the load determine the voltage.
Remember, the load has its own I-V characteristic that it must obey. For example, a resistive load obeys Ohm's Law \$V=IR\$.
What a 3-V 60-mA CC/CV supply does is force 3 V, unless doing so would require more than 60 mA, in which case it just provides 60 mA at whatever voltage (lower than 3 V) it takes to make that happen.
Assuming your LEDs' forward voltage is less than 3 V, your supply will operate as a 60-mA current supply, and the scenario will play out much as you have described it.
My scenario: 3v power supply directly from a battery.
A battery is more like a constant-voltage supply (with a series resistance), so this will indeed act differently than the constant-current scenario above.
Shouldn't the other two LEDs still draw 20mAh each?
Yes, they will continue to draw whatever it is they draw at 3 V. If they're the same LEDs as in the constant-current supply scenario (where we hypothesized that the forward voltage is lower than 3 V), then they will draw much more than 20 mA because you're driving them well above their rated forward voltage.
Also, as they heat or cool, their I-V curves change. So one might be okay running on a 3 V battery when first connected, but then start drawing more current as it warms up and eventually (or very quickly) burn itself out.
Boohoo, one LED breaks. That should be all. Why not?
With a constant voltage supply, one LED failing wouldn't affect the others.
With a real battery, you have to remember the internal resistance of the battery. If one LED fails, then that will tend to increase the actual output voltage of the battery after accounting for internal resistance, and that could cause the other LEDs to fail.
Best Answer
I don't know if I can recommend an exact LED to use (I'm constantly looking for better ones myself), but maybe I can point you in somewhat of the right direction.
Essentially, you want three things out of a bike light:
To start, your source voltage of 3.7 limits you to using 1 LED with current limiting resistor per parallel string. This isn't a bad thing, it just means you have to use more resistors as opposed to putting a few LEDs in series for double or triple that source voltage. The good news is that most lower-powered white LEDs have a forward voltage of 3 - 4V. I have to recommend that you DO use a resistor in series with each LED, even if the rated LED voltage is 3.7V. The LED current will fluctuate as it heats up and the LED forward voltage changes which can damage it or the battery. The resistor is there to make sure that current is limited.
You mentioned using 10mm LEDs rated for 25000mcd each @ 20mA. I'm guessing that the viewing angle on these is pretty low, such as 15 degrees. These types of LEDs are good for distance, but they don't work so well to light up things right in front of you. Here are some LEDs like that: 100 x 10mm 120000mcd @ 20mA, 12 degrees viewing angle - $40 (0.40 each). There are many LEDs for sale like this online.
If you are looking to use fewer LEDs that are higher power, something like this has a decent luminous intensity as well as a wide viewing angle: 62000mcd, 125 degrees - $12.95. The downside is that it will get very hot (needs the heat sink) and use quite a bit more current than your standard LEDs. This may be an even better choice: 300 lumens, 125 degrees - $6.95, but again, it will drain your battery a lot faster. However, pairing this type of LED with a focusing lens will reduce the viewing angle, but greatly increase the intensity. With the right lens, you can get everything you want out of one LED.
Consider the difference between a focused and unfocused flashlight: focusing the light beam increases the brightness, but less of the room is lit up; unfocusing the light reduces the brightness, but more of the room is lit up. To get the best light using the least amount of power, I would suggest you use a grouping of LEDs with a very high luminous intensity (mcd) paired with a group of LEDs with a high viewing angle (degrees). The easier route is to pair a high power LED with an appropriate lens, but this might not be ideal for you since you have such a small battery capacity. Either way will ensure you can see things far away, and the path in front of you is well lit.
Here is an online calculator which can estimate the output in lumens from the candella and viewing angle attributes: http://led.linear1.org/lumen.wiz
If you are trying to figure out your battery life, it can be estimated by dividing the amp-hour rating by the total current draw. This leaves you with the total hours of operation. Although, it isn't entirely accurate, and will vary based on the age and quality of the battery as well as the actual amount of current being drawn - some batteries do better with high discharge, some not so much.