Vf is called the "forward voltage" of the LED. What that means is that is that you need to apply a certain voltage across the LED to get a certain amount of current flow. For example: a Seoul P7 with specs of 3.7 Vf, 2800ma If, and 700 lumens would mean that you need to apply 3.7 volts across the LED to get a current flow through the LED of 2800mA and that would produce a typical light output of 700 lumens. If you applied less voltage, you would get less current and less lumens output. The only way to measure it is with the LED being supplied with power in a circuit. All you do is place your multimeter leads on either side of the LED and take a voltage reading. The Vf will change as the current flow changes and vice versa.
Vin and Vout are typically associated with driver circuits. When you are talking a simple circuit where you have a battery supplying a driver which in turn supplies an LED, Vin is the battery voltage and Vout is the Vf of the LED. Again, these are just measured by placing the multimeter leads across the appropriate terminals on the driver. That is why you need to have an idea of these values when buying components for a build, you need to make sure they are compatible. Normally, you will have an idea of what battery combination you want to use as a supply and what LED(s) you want drive. The battery voltage and current limits will let you know the Vin and Iin you can supply to the driver and the LED Vf and If specs will let you know the Vout and Iout you need the driver to supply.
If your battery voltage is higher than your needed LED voltage, you will need a buck (step down) type driver. If your battery voltage is lower than the needed LED voltage, you will need a boost (step up) type driver. There are a few buck/boost drivers but they are usually fairly expensive and also usually have a smaller operating range of voltages. You also need to take into account that the battery voltage will be relatively high fresh off the charger and will drop some once you put a load on it and then continue to drop further as the battery is depleted.
You can also do some theoretical calculations for a given setup. Lets say you have a set of six 2500mAH (milliamp hours) NiMH AA batteries (about 1.2V each under load) in a normal series connected battery pack. That will give you a total Vbatt of 7.2V and a capacity of 2500mAH. Lets assume you want to drive an LED with a 3.6Vf at 2800mA. We'll assume there are no losses in the driver circuit for the calculations, but in reality there are always losses so the actual results will be different. The total power, which is voltage times current, has to be the same going into and out of the driver. You can't get something for nothing. So if the battery is supplying twice the voltage needed by the LED, the battery current will be half the current needed by the LED. It is the driver that makes this conversion and controls the current to the LED to prevent burning it out. So in this example, you would expect the battery to be supplying about 1400 mA to the driver since the driver is supplying 2800mA to the LED. To figure runtime, you divide battery capacity by the current. 2500mAH / 1400mA = 1.78 hours or about an hour and 45 minutes. Fudging the numbers a little to allow for losses in the driver and reduced performance as the batteries are depleted would probably mean a real world runtime of about 80-90 minutes.
I forgot to mention, to measure current, you need to place the multimeter in series. The easy way to think of it is that you need to break a connection in the circuit and basically put the meter in place of a piece of the wire in a circuit.
