I'm not sure if you now fully understand it... maybe you do!
However I feel handicapped for I wished I had a nice video like the one you made to explain this!
So, let's have a look at your video together.
For some reason that escapes me, you make a difference between "with respect to" and "relative", although you don't really make use of that difference. Still your good word choice may be helpful for my explanation to you! I will consistently use "relative" for a difference in velocities (a vector subtraction), and "with respect to the ground" for the ground as reference system.
First you compare a velocity measurement of the bike by the ground with a velocity measurement of the ground by the bike.
That is called a "system transformation", which happens when you make a conversion of measurements of one reference system to those of another one. And the transformation that you use is called a "Galilean" transformation. Coincidentally, also in special relativity, v_bike-ground = -v_ground-bike; so you can indeed say "always". But that is really an exception.
I suppose that with v_bike-ground you do NOT intend the - as a minus sign, which would indicate a velocity difference (=relative velocity in some textbooks). You clearly say "velocity of the bike with respect to the ground", and "velocity of the bike as seen by the ground". Thus with that you take the ground as reference system. Perfect.
But then (2:20) you arrive at slippery ground. You claim that this is true for the velocity between any two objects: that velocity of B wrt A is
always the negative of the velocity of A wrt B. That is not necessarily true. It is where many textbooks go wrong; an object is not always a reference system. In fact it usually is not!
Take for example the moon. From the Earth you always see the same side of the moon. What is its motion with respect to the Earth? And what is the motion of the Earth with respect to the moon? An astronaut who lands on this side of the moon, will not see the Earth go down or rise.
Next you nicely introduce the concept of a "reference". Good!
My suggestion is that you say somewhere (after 4:00) that the rule that you are going to give is only true in classical mechanics. And if a student asks you why, I know that you will give the right answer.
Then, at 5:40 you introduce (but hardly noticeable!) a new concept: relative velocity.
You say that the relative velocity of the road and the limousine "cancel out". You probably mean that
with respect to the car, the velocity of the ground relative to the car is the negative of the velocity of the limousine relative to the ground, so that the velocity of the limousine relative to the car is zero. That is a straightforward vector subtraction, although in one dimension; it is all calculated using a single reference system, the car.
If that is indeed what you did, then you used there "relative velocity" definition 1 of my explanation in the link of my first reply. Do you copy that?
The advantage of that definition is that it is equally valid in special relativity. The advantage of the other definition (no.2) is that it looks somewhat simpler, even though it is in fact a system transformation. However, I noticed in discussions that it can hinder the capability to arrive at a good understanding of special relativity (which is why I nickname it "Newspeak"; you will know what I mean if you read Orwell!).
Using definition 1, I would say that the relative velocity of the limousine and the car are zero with respect to the car. And in such a case the relative velocity of the limousine and the car with respect to the ground is of course also zero.
.. but its an educational video! Cheers 
Vector addition of the same things always works. Therefore, a velocity difference in a single reference system is a vector subtraction that always works. That is different from a system transformation. 
