FMCW radar: Linear phase change determination with sinusoidal decomposition
Your understanding is mostly correct but no need of non-sinusoidal decomposition and all.


  Am I right in assuming that a constant velocity results in a LINEAR phase shift of the frequency component across chirps?


Yes, the change in phase due to time change $\Delta \tau$ is $\Delta\phi = 2\pi F (2\Delta \tau)$, $2\Delta\tau$ because signal has to travel back so double the distance covered by object. $F_c$ is the chirp frequency. $\Delta \phi = 4\pi F\frac{\Delta d}{c}$, where $\Delta d$ is the distance increment due to velocity of object and $c$ is velocity of light. Since wavelength $\lambda=F/c$, and $\Delta d=vT_c$ (where $T_c$ is chirp duration), phase change between 2 chirps is   $\Delta \phi = 4\pi \frac{vT_c}{\lambda}$.  So you can see that relative phase of peak bin of FFT output is linearly changing with respect to velocity $v$.


  If the phase shift is linear, how can FFT along a column determine the same?


Once you have identified the column corresponding to peak bin, take the FFT of phases of these peak bins column wise. If phase changes linearly with time, the FFT of phases would show up as a peak bin when you take FFT along the column. 

From the previous formula, we have $\Delta \phi$ as the change in phase between two consecutive chirps. So $\Delta \phi$ can be denoted as angular velocity $\omega$. Hence $v = \frac{\lambda \omega}{4\pi T_c}$. So if you take FFT of phases, if you see a peak at $\omega=\omega_0$, the velocity is $\frac{\lambda \omega_0}{4\pi T_c}$. If you see 2 distinct peaks, implies there are 2 equidistant objects approaching or retreating at different velocities at that point of time.