The important thing is the rise time (not the pulse repetition rate) of the signal, compared to the length of the trace.
If the signal can make several round trips of the line between the driver and receiver during the rise time of the signal, then we can ignore the transmission line effects. With a trace 200 mm long, which is about 1 ns electrical length assuming typical construction, a rise time of several nanoseconds will be slow enough to work unterminated. A sub-ns risetime will certainly cause problems unless the trace is properly terminated.
The easiest way to see what's going on is to use a simulator. This is the circuit I'm going to simulate. A 5 V step with a 10 nS risetime feeds a 100 Ω transmission line. The series termination resistors will either be 10 Ω for a mismatched driver, or 110 Ω (more or less matched, enough mismatch left to see what's going on). The shunt termination is either absent, or a nearly matched at 110 Ω.
simulate this circuit – Schematic created using CircuitLab
Let's start off with the ideal case, with shunt termination, below. The shun resistor is 110 Ω, the series resistor is 10 Ω, to represent a finite driver output impedance. This is expensive in terms of drive power, as the driver has to drive the full impedance of the line with the step, and the termination resistor at DC.
The line is 40 ns long, which means the input step has made its full swing well before any reflections return.
You can see the effect of the small mismatch as the reflections return, but they only produce a small ripple on the final waveform. The switching waveform is ideal at all points on the transmission line.
Now let's use a cheaper form of termination, series, below. The series resistor is 110 Ω with the shunt open circuit. The driver only has to drive 210 Ω with the step, and no DC driver power.
We only have a clean waveform at the end of the line. The start and midpoints of the line go up to 2.5 V initially, due to the voltage division between the series resistor and the line impedance. They stay there until the reflection returns from the end of the line and lifts the voltage to the full 5 V. If we had logic gates connected to those points, especially clock inputs, they could oscillate. Series termination can only be used to drive a single receiver at the end of the line.
What happens if we don't terminate a line this long? The series resistor is 10 Ω, a fairly strong driver with no attempt at matching, below.
Without the voltage division of the series resistor, the line goes up to more or less the full voltage at once. However, when the reflection returns, it now boosts the voltage to double, which will cause substrate diodes to conduct at the inputs to the gates. These are only designed to protect the inputs from EMI, and current through them could disturb normal operation, possibly even latchup.
Even worse, when the next reflection occurs, the voltage dips below 2.5 V, meaning a clock input will see a second edge. As time goes on, the reflections subside, energy gradually being absorbed in the driver output resistance. At some point, the reflections will stop switching any clock inputs on the line.
Finally, let's have a look at a short line, below. It's still unterminated, with no shunt resistor and a 10 Ω series resistor. The input step risetime is still 10 ns, but the line has been shortened to 2 ns, roughly 16" or 400 mm of track on a board.
When the reflection gets back to the source end of the line, the source voltage has not risen very far, and the reflected signal is still quite small. Although you can see the reflections do influence the trajectory of the waveform, the signal is still 'clean enough'. There are no extra transitions crossing 2.5 V. The ringing at the top of the waveform will probably not be turning on any substrate diodes in the receiver.
At some point between 2 ns and 40 ns, the waveform will breach some threshold of acceptability. Perhaps >1 V overshoot? Perhaps the leading edge of the voltage waveform becoming non-monotonic? Perhaps the waveform dipping below the switching threshold? Any particular situation might have its own criterion for successful operation. But well away from the threshold, we can easily see what we mean by 'short enough to be OK', and 'long enough to give a problem'.