You're getting the expected result. What you see is the normal behavior of diodes in series and it's completely normal to have one resistor and a string of LEDs connected after it.
What's basically happening is this: When they told you that the forward voltage is 2 V, they lied. It actually depends on the current going through the LED and you can consider the 2 V some sort of nominal value, but the exact drop should be read in the datasheet (if it's available).
In general case when you want to connect diodes in series, you use this formula for resistor:
$$ R= \frac {V_{supply}-NV_{f}}{I_{f}}$$
where the N is number of diodes you have.
This way it turns into simple Ohm's law. But in your case, you're approaching the border at which the above formula will not hold. You basically have a circuit with one branch only and the current going through that branch isn't going to much change with the number of LEDs if the voltage of the supply is high enough to be higher than LED forward voltage.
Take a look at this diagram from Wikipedia:
Notice the point marked \$ V_d\$. For this diode, once the voltage at the diode terminals reaches that point, the current will start quickly increasing with only a small change in voltage. That is why adding more LEDs doesn't immediately affect current. The voltage is high enough that all LEDs will conduct. Should you for example put 10 LEDs in series, the voltage will be too low and they will either show barely noticeable light levels or stay off.
Next, let's take a look at the different voltages you got at the LEDs. Again take a look at the curve for the diode from the Wikipedia. The \$V_d\$ point for each diode made is different and there are some tolerances here. So some diodes of same model number will at same current have a bit larger voltage drop and others will have a bit smaller voltage drop.
Next about LEDs in series. There is nothing wrong with that, but you're still not doing it right. Using the formula I provided, you should set the resistor so that the LEDs will be within their rated current. If you fulfill that condition, there's absolutely nothing wrong with having multiple LEDs connected in series, should you have voltage to spare.
One thing you need to remember is that voltage is relative. Voltage is a potential difference and it makes no sense to discuss voltage without a 'zero' reference.
In the case of your LED circuit there will be a voltage across the battery, a voltage across the LED, and a voltage across the resistor. If you add up all the voltages as you go around the loop, you get zero – up 9 at the battery, down 6 at the resistor, down 3 at the LED, the total is zero and you're back at the same point in the circuit. The LED only sees the difference in voltage between its two leads, as does the resistor. Since only the difference is important it makes no difference what order the parts are connected in.
As for current, it is only the same along a continuous path. Electrons are not created or destroyed (what goes in must come out). Since there is only one possible path for the electrons to take in your circuit, the current will be the same through all of the components. In a parallel circuit you see the opposite: all of the components have the same voltage, but the currents will be different.
Best Answer
Switching the positions of the LED and resistor will have no effect on the brightness of the LED because they would still be in series. The current through them would be exactly the same. Thus the brightness of the LED would be the same. The voltage drops across the resistor and the LED would be the same since the current through them is the same. This is a characteristic of all series circuits.