Continuous varying impedances are used all the time for impedance matching. If you have a very capacitive part of a trace (for example, where a large component pad might be), you can have a relatively inductive transition before or after it to "balance" it out.
What will end up happening is that the reflections will "stack up" but, instead of being at one point (a VSWR peak), it will be moderately spread out. You can still imagine it discretely, but in small steps.
And also remember, if you have a small reflection point, any backward reflection after THAT will be reflected slightly FORWARD, and so on.
Anyway, the good gents at http://www.microwaves101.com/encyclopedia/klopfenstein.cfm always have a nice, in depth explanation.
edit: I didn't completely answer your question. "How it would look" is dependent a bit on how you are describing it. In the frequency domain, what you'll probably get is a VSWR that is "de-Q'd". You'll go from a nice sharp peak at midband to a more gradual, broader band response.
In the time domain....well, I don't work with the time domain as much but I would imagine you would have a lower amplitude, longer pulsewidth "ringing" or reflection.
What do the red and green arrows represent?
Green is the forward travelling wave. Red is the reverse travelling wave.
The length of the arrow indicates the amplitude, and its angle of rotation around the axis represents its phase at each point along the line.
From the way they relate to each other at the terminations, we can tell the arrows must represent the voltage wave rather than the current wave.
Why do they have those orientations?
I suspect the orientation of the arrows relative to the direction of propagation is meant to represent the phase of each wave (forward and reverse travelling) at each point.
In 3), where there is little reflection, is the total voltage wave a standing wave?
The standing wave is the superposition of the forward and reverse travelling waves.
In 4), why is the red line completely flat?
Because there is only the forward travelling wave, its amplitude is the same at all points along the line.
In 4), why only green arrows?
Because red arrows represent the reverse travelling wave, and since the termination is a perfect match there's no reverse travelling wave.
What are the impedance's values Z1,...,Z7 in each picture?
I suspect they are the impedance looking into the line at each of the points labelled Z1, Z2, etc., in the diagrams.
Okay, I want to go back to one of your assumptions before you enumerated your specific questions,
In this case there is total reflection, and the total voltage wave is a standing wave and, if I understood correctly, it is that in red.
If I have interpreted it correctly, red represents the reverse travelling wave (which I deduce because there are no red arrows when the termination is well matched).
The standing wave is the superposition of the forward and reverse travelling waves, not the reverse travelling wave on its own.
Best Answer
We'll start with what a square wave is.
A square wave is a bunch of sine waves added together. You have a primary, and a bunch of harmonics added to it. The harmonics are in a particular phase relationship to the primary. The sum of all the waves is a square wave. If you leave any of the sine waves out (or change their phase,) then the sum looks less like an ideal sine wave.
This image from the Wikipedia page illustrates it clearly:
Now, on to the reflections.
If your transmission line were terminated in a perfect resistor that doesn't match the line impedance (and there were no effects from capacitance or inductance) then the reflection would be a nice, sharp copy of your original square wave.
In any real circuit, you will have capacitances and inductances to deal with.
The inductance, capacitance, and resistance of your transmission line will interact with the inductance, capacitance, and resistance of the termination. The result is an impedance that varies with frequency.
A square wave is a collection of sine waves of different frequencies, so different parts of it are more or less strongly reflected. Some frequencies will be slightly delayed, and some won't.
The result is often that some part is reflected as a somewhat proper looking square wave while other parts are reflected differently enough that you can see the sine wave itself.
When neither end is properly terminated, then each sine in the square wave is reflected depending where exactly it "hits the end."
Say you have a transmission line 1 meter long, and a wave one meter long. In this case, there will be no reflection - the wave fits exactly in the transmission line.
Now keep the 1 meter transmission line, but change to a wave that is 67 centimeters long. The wave doesn't fit exactly in the transmission line anymore. Part of it will be reflected.
Put the one meter wave and the 67 centimeter wave into the same transmission line at the same time, and you will only see reflections from the 67 centimeter wave.
Now consider your square wave (made of waves of different lengths.) Some parts will reflect more strongly than other parts. If most of the sine waves reflect together, then you might get something that looks like a square wave. You might also get a situation where only a few (or one) frequency reflects strongly - in which case you get something that looks like a sine wave as a reflection. That's ringing.