It's a simple matter of definitions. In either direction, there is a voltage above which the diode begins to conduct a large current for a small increase (or decrease in the reverse case) in voltage. The finer details of the current-voltage function in each direction are somewhat different, but as a first order approximation, above a minimum (reverse breakdown voltage) and below a maximum (forward voltage), a diode does not conduct at all, and at voltages below or above these limits, it conducts a lot. This approximation is sufficient for most engineering purposes.
The reason for the difference in terms is that the underlying physical mechanism is quite different. The forward voltage has to do with the nature of the semiconductor, and for all silicon PN diodes, this will be in the neighborhood of 0.65V. The reverse breakdown voltage additionally depends on the geometry and design of the device, and quite a range of values are attainable, even among silicon PN diodes.
Also, don't let the term "breakdown" suggest that the diode "breaks". What is "breaking down" is the usual state of the diode that prevents reverse current flow. Once the reverse breakdown voltage is exceeded, the diode isn't necessarily damaged. However, a large current will flow, and if this current isn't limited (say, by a series resistor), then the diode will overheat. Then it will be damaged.
Note this isn't really any different from the case when the diode is forward biased. Any attempt to apply significantly more than the forward voltage will result in a very large current which overheats the diode and destroys it. Limiting the current avoids damage.
Ordinary silicon diodes (example, 1n4148) are not often intentionally operated in reverse breakdown. Their behavior in this mode of operation is not usually specified except for some minimum reverse breakdown voltage. There are other diodes, such as Zener diodes, which are usually operated in reverse breakdown (though the physical mechanism is somewhat different). These diodes have more completely specified behavior in this operation, because by virtue of their design, the relevant operational parameters can be more predictable and stable.
You can't use Ohm's law to calculate the current through a diode.
However, the resistor is in series with the diode and, as you've learned, series connected circuit elements have identical currents.
Thus, you can use Ohm's Law to find the current through the resistor and diode but, to do that, you need the voltage across the resistor.
When you subtract the diode voltage from the source voltage, by KVL, what remains is the voltage across the resistor \$V_R\$.
Now, if you know the desired current, e.g., \$20mA\$, you can calculate the resistance required for that series current.
Thus, by Ohm's Law, the desired resistor value is:
$$R = \dfrac{V_R}{20mA} = \dfrac{V_S - V_D}{20mA} = \dfrac{9V - 1.7V}{20mA}$$
Best Answer
Diode is not the problem. You can remove it and replace it with a piece of wire to see that for yourself. Make a simpler circuit work first. Then when you add the diode it'll keep on working (as long as there's enough voltage driving the circuit).
You assumed that the diode is a problem, whereas before such assumption you should ask yourself: do you know enough to know that the diode is a problem? Remember that the voltage drop across a diode is ~logarithm of the current, so all you're seeing here is a diode acting like a rather sensitive current meter and telling you that microamps are flowing through it. You can replace the diode with a microammeter and see what the current is. I'd expect anywhere between 0.1uA and 10uA.
Once you get the circuit working, you should play with that diode - put it in series with a resistor, measure voltage across the resistor to determine the current, then measure the voltage across the diode. Use various resistors to "sweep" the current across the measurement range. Plot the diode voltage vs. current over at least 4 order of magnitude - say between 0.1mA and 100mA, and see whether the response is logarithmic (after all, I can be talking nonsense - you really should see it for yourself since it's easy and thus no reason at all to blindly trust me).