Why CW radar I-Q signals do not contain Doppler frequency component?
I've been reading your comments and I see your understanding has grown a little.

Your transmit signal is set to some transmit frequency and reference phase: $$f_t\ and\ \phi$$ the phase could be anything but let's say it's zero for our purposes. A simple pulse-Doppler radar will transmit a pulse of the form: $$\cos(2{\pi}f_t{t})$$

Your return signal from a moving target will contain a frequency shifted version with an added phase: $$\cos(2{\pi}(f_t + f_D)t + \phi(t))$$

The signal will come to the mixer that will split and then mix the split signals individually with: $$\cos(2{\pi}f_t{t}) \ \ \sin(2{\pi}f_t{t}) $$

Which produces your I and Q channels (cosine and sine, respectively). The resulting signals after low-pass filtering will remove the transmit frequency and you'll end up with:  $$\cos(2{\pi}f_D{t} + \phi(t)) \ and\ \sin(2{\pi}f_D{t} + \phi(t))$$

At this point, the information you want is available. You can pass either I or Q through a Doppler filter bank to determine the Doppler frequency of the signal. Doing a Fast Fourier Transform (FFT) will give you the spectrum of the received signal which will peak at the Doppler frequency received. The phase of the signal at this point is not as important.

Finding the magnitude from I and Q is for just that, finding the magnitude of the returned signal mainly for detection and display purposes.

This is a very simple overview and Doppler radar demodulation contains more steps that are not shown here, but this should give you a good idea.