For a given set of control points $b_0, b_1, \ldots, b_n$, the Bezier curve is defined as $$b^n(t) := \sum_{j=0}^n b_j B_j^n(t),$$ where $B_j^n(t):=\binom{n}{j}t^i(1-t)^{n-i}$ are Bernstein polynomials. Let's assume that the curve is defined on the closed unit interval $[0,1]$; if not, we can transform $a\le u\le b$ with $t := (u-a)/(b-a)$.
The derivative of the Bezier curve is $\frac{\partial}{\partial t}b^n(t) = n\sum_{j=0}^{n-1}\Delta b_j B_j^{n-1}(t),$ where $\Delta b_n := b_n-b_{n-1}$, which is again a Bezier curve.
So the derivative of a Bezier curve is a Bezier curve. Is there an interesting geometric interpretation of the derivative of the given Bezier curve? (For example, the value of the derivative at a point of a real function in one variable is the slope of the tangent at that point.)
I don't know if it's interesting, but here is how you can interpret the facts you cited.
The first thing, of course, is that the derivative vector $b'(t)$ is in the direction of the tangent vector at parameter value $t$. But this is true of any parametric curve, not just Bezier curves.
Take the $n$ vectors $\Delta b_n$ and relocate them so that their "tails" are at the origin. The "heads" of these vectors form a polygon (a polyline, really) that can be used as the control polygon of a Bezier curve $g$ of degree $n-1$. As you noted, the vector $g(t)$ is the derivative vector of the original curve at the parameter value $t$ (with some scaling). The curve $g$ is called the hodograph of the original curve. Bezier's original writings contain quite a bit of discussion of hodographs. Hodographs are used in other fields, too, outside of CAGD.
One thing that's useful: the usual Bezier convex hull property applies to the curve $g$, and you can use this to obtain bounds on the derivative vector of the original curve. Continuing with further differentiation steps, you can obtain bounds on the second, third, and successive derivatives. Bounds on derivatives are often needed when doing approximation, so this technique is useful.