Misunderstanding in partial derivative of a parametric surface

[woeful apex]

I've watched at least 3 times from 2:37 to 3:39 of https://www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/partial-derivatives-of-vector-valued-functions/v/partial-derivative-of-a-parametric-surface-part-2 (notice the transcribe tab under the player) and I don't get it. AFAIK he start by drawing ds then drawing the corresponding change in the same color, which I understand that far; he then draws something else in another color and I don't understand what that correspond to in the overall goal. It seems to me that the heart of the issue is

3:19 So the result that you get
3:22 is a tangent vector that's not puny, not a tiny nudge,
3:26 but is actually a sizable tangent vector.
but that's just my flawed understanding of that section of the video. Help?

[coral sinew]

the first color is dv, which is tiny (as small as possible)

once you divide it by ds, you get dv/ds (which is no longer small). It becomes a rate of 2 infinitesimally small changes

since it's a rate. it's no longer "as small as possible"

that gives you the partial derivative.

Get it ?

[woeful apex]

Did you notice the 1st color is drawn 2X, before and after the transformation? Isn't the 1st drawing dv, then the 2nd drawing (of the same color) dv/dt?

[coral sinew]

So here are the steps :

1. The input is a tiny nudge in the s direction (ds)
2. The output is a tiny nudge in vector v (dv)
3. We are interested in the ratio (dv/ds), which is the partial derivative of v by s

Is it clearer ?