In Kleppner's "intuitive" explanation of acceleration equations in polar coordinates he uses a geometric argument based on the figures on the left, but I don't get how the angles between the velocities are the same as the angle between the position vectors. I find it simple when the curve is a circle, but in the general case I have no idea how to prove it.
This is a pure case of geometry. In case you miss to see it: just lift both the velocities off and superimpose them with their corresponding position vectors, you will see that the angles are same.
The angle that the position vector makes with the velocity vector is pi/2 radians. The velocities are tangential to the path at any point t in the trajectory, so the angle between the velocities at two different points is also the same(two straight lines make the same angle with respect to each other as that the lines normal to them make).
Let $u$ denote a time-dependent vector. Working at second order in $\mathrm dt$ throughout, we'll use the identities$$\begin{align}\sqrt{1+A\mathrm dt+B\mathrm dt^2}&=1+\frac12A\mathrm dt+\left(\frac12B-\frac18A^2\right)\mathrm dt^2,\\\frac{1+A\mathrm dt+B\mathrm dt^2}{1+C\mathrm dt+D\mathrm dt^2}&=1+(A-C)\mathrm dt+(B-D+C(C-A))\mathrm dt^2.\end{align}$$ In the calculations below, all quantities are at time $t$ unless indicated otherwise:$$\begin{align}u\cdot u(t+\mathrm dt)&=u\cdot(u+\dot{u}\mathrm dt+\tfrac12\ddot{u}\mathrm dt^2)\\&=u^2\left(1+\frac{u\cdot\dot{u}}{u^2}\mathrm dt+\frac{u\cdot\ddot{u}}{2u^2}\mathrm dt^2\right),\\|u(t+dt)|^2&=(u+\dot{u}\mathrm dt+\tfrac12\ddot{u}\mathrm dt^2)\cdot(u+\dot{u}\mathrm dt+\tfrac12\ddot{u}\mathrm dt^2)\\&=u^2\left(1+\frac{2u\cdot\dot{u}}{u^2}\mathrm dt+\frac{u\cdot\ddot{u}+\dot{u}^2}{u^2}\mathrm dt^2\right),\\|u(t+\mathrm dt)|&=|u|\left(1+\frac{u\cdot\dot{u}}{u^2}\mathrm dt+\frac{u\cdot\ddot{u}+\dot{u}^2-(u\cdot\dot{u})^2/u^2}{2u^2}\mathrm dt^2\right),\end{align}$$so the cosine of the angle between $u,\,u(t+\mathrm dt)$ is$$\frac{1+\frac{u\cdot\dot{u}}{u^2}\mathrm dt+\frac{u\cdot\ddot{u}}{2u^2}\mathrm dt^2}{1+\frac{u\cdot\dot{u}}{u^2}\mathrm dt+\frac{u\cdot\ddot{u}+\dot{u}^2-(u\cdot\dot{u})^2/u^2}{2u^2}\mathrm dt^2}=1-\frac{u^2\dot{u}^2-(u\cdot\dot{u})^2}{u^4}\mathrm dt^2,$$making the angle $\varpi(u)\mathrm dt$ with${}^\dagger$ $$\varpi(u):=u^{-2}\sqrt{2\left(u^2\dot{u}^2-(u\cdot\dot{u})^2\right)}$$(because $\cos(\varpi\mathrm dt)=1-\tfrac12\varpi^2\mathrm dt^2$). Since the force is radial, $a=\Omega^2r$ with $\Omega\ge0$ an in general time-dependent scalar of dimension $\mathsf{T^{-1}}$, so$$\varpi(v)=v^{-2}\sqrt{2(v^2\Omega^4r^2-(\Omega^2v\cdot r)^2)}=\frac{\Omega^2r^2}{v^2}\varpi(r)=\frac{a\cdot r}{v^2}\varpi(r).$$By the virial theorem $v^2=a\cdot r$, so $\varpi(v)=\varpi(r)$.
${}^\dagger$ The little-known symbol $\varpi$ is a $\pi$ with the legs' feet folded inward to meet, but its resemblance to $\omega$ makes it useful here as a $\mathrm dt$ coefficient in an angle. Don't confuse it with the angular velocity $\dot{\theta}$ usually denoted $\omega$.
The "tangential" velocities are perpendicular to the position vectors. There is a basic theorem saying that angles with perpendicular sides are equal (or supplementary). https://etc.usf.edu/clipart/70000/70087/70087_anglesum.htm
But you must be careful about the terms here. The velocity vector is always tangent to the trajectory. The "tangential" component means the components along the unit vector $ \hat{\theta} $ in polar coordinates. It is perpendicular to the unit vector $ \hat{r} $ which is along the position vector $ \vec{r} $. The figure does not show the tangential components as perpendicular to $ \vec{r} $. The dashed line in the figure cannot be the trajectory. I don't understand what that line is supposed to be. The first figure looks wrong to me.
