Sorry I should have used latex, but the editor doesn't work in my browser (chrome).
Well, it is pointing in the right direction and it has a magnitude of unity. That pretty much defines a unit vector.
I think another one of your attachments, from another thread, actually made this explicit.
Just look at it intuitively instead. The Δs is the small distance between two close points. The derivatives dx/ds, dy/ds and dz/ds represent the rate of change of x, y and z relative to changes in s, as you move along the line that connects the two close points.
Let's say s changes by 1 micro-unit a the infinitesimal level. Then, the corresponding changes in x, y and z (i.e. dx, dy and dz) will be rectangular components of the offset in s (i.e. ds), and the vector dr=dx i + dy j + dz k will have length of 1 micro-unit also. If you draw this out, you will see that the line segment between s and s+ds is the diagonal of a box with lengths dx, dy and dz. Hence, the magnitude of the vector dr=dx i + dy j +dz k , must be equal to ds since distance across a box is given by the Pythagorean Theorem and this formula is identical to vector magnitude.
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