Forget about breaking vectors into a scalar-unit product. You don't need it. You take a derivative of a vector exactly the same way you do of scalar quantity. If you have to actually work out components, you have to take it all apart, but in this derivation, just pretend they are scalars (except for cross products).
Any time I have a d/dt of ANYTHING, I apply the rule I first stated in this thread. It becomes a cross product with omega plus a partial derivative. So let me take one particular step apart, and hopefully you'll see how the rest fall in place. I need to take d/dt of the r vector twice. So I start by taking it once and have the following expression.
\frac{d}{dt}\left( \vec{\omega} \times \vec{r} + \frac{\partial \vec{r}}{\partial t}\right)
That's a derivative of the sum, which is a sum of derivatives.
\left(\frac{d}{dt}\left(\vec{\omega} \times \vec{r}\right) \right) + \left(\frac{d}{dt}\frac{\partial \vec{r}}{\partial t}\right)
Now let me continue with just the first term in the sum above. Just like with ordinary products, d/dt(axb) = (da/dt)xb + ax(db/dt) for any vector quantities a and b. So here is what I get.
\frac{d}{dt}\left(\vec{\omega} \times \vec{r}\right) = \left(\frac{d}{dt}\vec{\omega}\right) \times \vec{r} + \vec{\omega} \times \left(\frac{d}{dt}\vec{r}\right)
So now I have d/dt acting on vectors, which again I can decompose using the rule from the above. Again, let me just take the first term of this sum.
\left(\frac{d}{dt}\vec{\omega}\right) \times \vec{r} = \left(\vec{\omega} \times \vec{\omega} + \frac{\partial \vec{\omega}}{\partial t}\right) \times \vec{r}
As you correctly pointed out, ωxω term is zero. So only one term from this makes it to the final result.
\left( \frac{\partial \vec{\omega}}{\partial t}\right) \times \vec{r}
What does it mean? Let me denote partial derivatives with respect to time like this, ω'. For simplicity, let's assume ω' has the same direction as ω. That means that only the magnitude of ω changed, and not the direction. The direction of this term will be same as direction of ωxr, that is, perpendicular to the r and ω. Or in other words, along the rotation. So all that term tells you is that if you are standing inside a rotating space station, and it suddenly starts to rotate further, you'll get thrown against the rotation.
This is the only term that depends on changes in ω. All others depend either on changes in r or just on location. Let's look at them.
ω'xr - we just discussed.
ωx(ωxr) - centrifugal term. That should be pretty clear.
2ωxr' - Coriolis effect. It's easiest to understand if you simply think about r moving towards or away from the center. Then r' is in the same direction as r, and ωxr' is again aligned with direction of rotation. It just says that if you move towards center, you'll feel as if you are pulled along with rotation, and if you are moving away, you are pulled against the rotation. (Remember, fictitious force acting on body at r is -m(d²r/dt²))
r'' - This is a term from the r actually accelerating relative to local frame. If you are on the rotating station, along with all other effects, if you simply decide to walk around, you will fee effects of your own acceleration. That's what this term tells you.
I hope that clears up most of the problems. If anything is still not clear, let me know, and I'll try to get into more detail. If you have problems with full vs partial derivatives, we can talk about that too, because this is a good illustration.
Do you think about it logically only or do you have some method to ensure you get all terms right?
I do try to picture what the situation actually looks like, and what forces are going to be in place. Sometimes it helps to picture yourself in the situation, because your body is pretty good at predicting accelerations. But you cannot completely rely on this. Some of the effects are not obvious. For example, if you start running against rotation of a station, you get lighter. That's not immediately apparent, but it is accounted for in the Coriolis term. It does become clear in the extreme case, though. If you are moving against rotation at the same rate that the station rotates, you are no longer rotating with the station, and so no longer experience centrifugal effect, but the centrifugal term is still there. The Coriolis term simply cancels it.
So you can use these sort of things to check your work, and look for suspicious terms, but don't over-rely on it, and make sure you derive everything step-by-step.
