Why does the tangent vector measure the rate of change of the angle which neighboring tangent make with the tangent? I'm reading the classical Manfredo's differential geometry book and I couldn't prove formally the following statement written on page 16:

Why does the tangent vector measure the rate of change of the angle which neighboring tangent make with the tangent at $s$?
 A: For simplicity, lets explain this at the point where $s=0$.
Let's try to do this the naive way first, and we will see where we get stuck.
Attempt 1 The angle formed by the tangent vector to the curve at time $0$ and the tangent vector to the curve at time $s$ is the angle $\theta(s)$ formed by the two vectors $\alpha'(0)$ and $\alpha'(s)$.  Since these are unit vectors, this angle is given by the formula 
$$\cos(\theta(s)) = \alpha'(0)\cdot \alpha'(s)$$
Note that when $s=0$, the RHS is $|\alpha'(0)|^2 =1$, so $\theta(0) = 0$.
If we evaluate the derivative of both sides of this equation at time $s=0$, we get 
$$-\sin(\theta(0)) \theta'(0) = \frac{d}{ds}\bigg\vert_{s=0}\alpha'(0)\cdot \alpha'(s) = \alpha'(0)\cdot \left(\frac{d}{ds}\bigg\vert_{s=0}\alpha'(s)\right) = \alpha'(0)\cdot \alpha''(0)$$
(since dot-product is linear in the second argument). Unfortunately, this equation tells us nothing about $\theta'(0)$ since $\sin(0)= 0$ (indeed, this tells us that $\alpha'(0)\cdot \alpha''(0)=0 $ always, which is why the equation won't tell us $\theta'(0)$).
Ok, so the naive solution doesn't work, but the problem was differentiating $\cos$ to $\sin$, so we have an idea.
Attempt 2 Let's use a different formula relating $\theta$ and the unit vectors $\alpha'(0)$ and $\alpha'(s)$:
$$\sin(\theta(s)) = |\alpha'(0)\times \alpha'(s)|$$
If we differentiate both sides of this equation away from $s=0$, we get 
$$\cos(\theta(s)) \theta'(s) = \frac{d}{ds}|\alpha'(0)\times \alpha'(s)| = \frac{\alpha'(0)\times \alpha''(s)}{2|\alpha'(0)\times \alpha'(s)|}$$
which is the same as 
$$2\sin(\theta(s))\cos(\theta(s)) \theta'(s) = \alpha'(0)\times \alpha''(s)$$
and I'm still stuck. Huh.
Attempt 3 Ok, let's finish this answer incorporating the argument I linked in the comments. Differentiating the formula from Attempt 1 a second time, we get that 
$$-\cos(\theta(s)) (\theta'(s))^2 -\sin(\theta(s))\theta''(s) = \alpha'(0)\cdot \alpha'''(s)$$
Plugging in $s=0$ yields
$$\theta'(0)^2 = - \alpha'(0)\cdot \alpha'''(0).$$
But we also know that 
$$0 = \frac{d}{ds}\alpha'(s)\cdot \alpha''(s) = \alpha''(s)\cdot \alpha''(s)+\alpha'(s)\cdot \alpha'''(s) $$
so 
$$\theta'(0)^2 = \alpha''(0)\cdot \alpha''(0) = |\alpha''(0)|^2. $$
