Help with caption from Kittel: Can we get phase angle difference from dot product?

Generally, the local phase angle $alpha$ of a wave described by $$psi=e^{ivec kcdot vec r}=e^{ialpha(vec r)}$$ is of course $$alpha(vec r)=vec kcdot vec r$$ That is just the very definition of phase $alpha$ of a plane wave (in stationary description, i.e. temporal phase $e^{-iomega t}$ has been divided out of the wave).

Due to that, all the wave fronts (the lines that look like "stair steps" perpendicular to the peripheral rays of the beam in Kittel's figure) have the same phase. In mathematical terms: $$vec kcdot vec r=const.$$ defines a plane (or in this 2D figure, a line).

From that it is clear that the phase difference between $r_O$ and $r_{dV}$ is given by the quotient of the respective phase factors: $$e^{ideltaalpha} = frac{e^{ivec k cdot vec r_{dV}}}{e^{ivec k cdot vec r_O}}=e^{ivec k cdot (vec r_{dV}-vec r_O)}=e^{ivec k cdot vec r}$$ and hence the phase difference is $$deltaalpha=vec k cdot vec r$$ Again, this is just the definition of the term "phase difference".

As to your second question: suppose that $chi$ was the angle opposite to $varphi$ next to the other leg of the right (in the sense of orthogonal) triangle. Then $chi$ is the angle that corresponds to the dot product: $$vec kcdot vec r = krcoschi=krcos (pi/2-varphi)=krsin varphi$$ Since the angular wave number is defined by $$k=2pi/lambda$$ (you want to have a phase difference of $2pi$ if $r=lambda$ in $e^{ikr}$, i.e. exactly one period of the wave after a distance of one wavelength), you finally get $$vec kcdot vec r=frac{2pi r}{lambda}sin varphi$$ By the way: the same formulae are also useful for single or double slit diffraction.