In the classical Doppler effect, you should think of $v_s$ and $v_r$ to be the velocities of the source and receiver relative to the medium, at the time when the signal was sent and received. In an accelerated reference frame, it is essentially the medium that is moving. Remember that the Doppler shift concerns the frequency sent by the source compared to the frequency seen by the receiver. So it does not depend on which reference frame you, as a third party, is in, but only depends on the intrinsic wave speed of the medium $v$ (which we assume to be constant, when measured in a frame in which the medium is at rest), and the velocities relative to the medium of the source and receivers.
With that said, that means you can compute relative to the unaccelerated reference frame by doing a coordinate change.
Suppose you are in a reference frame in which the medium itself is moving with variable velocity $\tilde{v}(t)$, and you measure the source and receivers to be moving with velocity $\tilde{v}_s(t)$ and $\tilde{v}_r(t)$. Then the Doppler effect can be computed by looking at the reference frame in which the medium is not moving (so the corresponding velocities are $v_s(t) = \tilde{v}_s(t) - \tilde{v}(t)$ and similarly $\tilde{v}_r(t)$). If in this reference frame $v_s$ and $v_r$ are not constant in time, then given a time $t_r$ when the signal was received, you will have to compute (using that the wave travels with constant speed $v$ and using the trajectory of the receiver versus the source) the time $t_s$ at which the signal was sent, and plug those values into your Doppler shift expression (with $v_s \to v_s(t_s)$ and $v_r \to v_r(t_r)$).
If the wave speed itself $v$ depends on the macroscopic of the entire medium, then you will have to solve the very complicated system for elastodynamics/acoustic in the medium, for which the final answer would depend on how you model the medium.
Edit for the new clarification
In the particular case, assume you, the receiver at the platform is measuring the signal at time $t_0 = 0$. And choose coordinate system so that you are at the origin. Let $x(t)$ denote the position of the train that is approaching the station. (or leaving the station, you just need to adjust the sign appropriately.) The signal you observe will have been emitted by the train at time $t_s < 0$, when $|x(t_s)| = v \cdot (0-t_s)$ where $v$ is the speed of sound in air. (For the time being we'll assume you have a non-supersonic train, so there is a unique time $t_s$ that satisfies the requirements.) Then the expression for the doppler shift would be
$$ f = \left( \frac{v}{v + \dot{x}(t_s)}\right) f_0 $$
Unfortunately, solving for $t_s$ is generally hard. But we can get a good estimate on $f$ if we assume that our train is reasonably close to the station when you are making the observation, and if the train is not accelerating too much, by using Newton's method.
To the zero'th order you'd get $t_s = 0$. To the first order in Newton's method, you can solve and get $t_s ~ x(0) / (v - \dot{x}(0))$. Then we can approximate the velocity $\dot{x}(t_s) \sim \dot{x}(0) + \ddot{x}(0) t_s$ which leads to a first order correction to the doppler shift formula
$$ f \sim \left( \frac{v}{v+ \dot{x}(0) + \frac{\ddot{x}(0) x(0)}{v - \dot{x}(0)}}\right) f_0 $$
Here's a similar question which will be helpful to resolve your apparent perception of Doppler effect. Because, Doppler effect is real.
So would it be right to say that when the wave is reflected from B, we can think of it as a source kept on the car B?
No. Because the frequency has already been altered (probably increased) by the moving car and the wind (which can be found using classical velocity addition) and finally, the high frequency sound has been returned to B, which reflects it back - which again gets altered by velocity of B and wind flowing in opposite direction.
Due to the speed of the wind, can we substitute the relative velocity of the source and the receiver in the formula?
You should use the velocity addition formula. Don't forget to account for wind too...
Does the wavelength of sound change if the medium is moving? In particular, if the wind is blowing at the speed of sound in the same direction as sound, what would be it's wavelength?
The speed of sound is a constant at a particular medium. So, if the frequency of sound changes, the wavelength should also change in order for $c$ to be constant. If the wind is blowing the opposite direction to the source of sound at $c$ (no relativistics...), the wavelength is $0$. Because, always keep in mind the sound is pushing & pushing & pushing of the molecules. If there's an opposition pushing it with the same velocity, then its whole energy would be spent trying to oppose wind and finally FAIL...
Best Answer
To answer your question, no it is not possible to derive the equations you call (1) and (2) from each other because $v_s$ and $v_r$ are independent of each other, and independent of the constants $f_0$ and $v_0$.
However, we can very easily either derive the equations separately and then combine them by multiplication, or simply derive them together. Let's do the latter:
Using your notation, let
$v_0$ = the speed of sound, positive in the direction from source to observer.
$v_r$ = the velocity of the observer, positive in the direction away from the source
$v_s$ = the velocity of the source, positive in the direction toward the observer
(Everything is positve from left to right, if the source is left of the observer)
$f_0$ = the source frequency
$f_r$ = the observed frequency
$\lambda_0$ = the source wavelength
We'll use the relation $$f_0=\frac{v_0}{\lambda_0},$$ but for an observer moving toward or away from the source, $v_0$ is replaced by $v'$:$$ v' = v-v_r$$
and for a moving source chasing or running away from its own waves, $\lambda_0$ is replaced by $\lambda'$ $$\lambda' = \frac{v-v_s}{f}$$ so $$f_r=\frac{v'}{\lambda'}=f_0(\frac{v-v_r}{v-v_s})$$ Note that for either a stationary observer or a stationary source, the above expression reduces to your equations (1) and (2), respectively (except that you didn't specify positive and negative directions). This is what dmckee meant in his comment above by saying that your equations were special cases.