However, I feel like having the denominator c+v is wrong, as in some
sense I'm adding velocity v to the speed of light which is not
allowed in SR.
It is permissible, in SR, to have 'non-physical speeds' in excess of c.
For example, you observe two trains speeding towards on another. You measure the speed of each train to be say, $0.9c$ relative to the track.
Then, according to you, their closing speed is $1.8c$, i.e., if at some time, the trains are separated by a distance $d$, the time to impact is
$$\frac{d}{0.9c + 0.9c}=\frac{d}{1.8c}$$
This is fine because there is no physical object with speed $1.8c$.
However, the speed of one train, according to an observer on the other train, must be less than $c$ and is given by the relativistic velocity addition formula.
So, the closing speeds, according to A, of B and the light beams are $c + v$ and $c - v$.
UPDATE to address the comments:
Can I just say that B's clocks are synchronized with A's clocks?
No. Regarding your second question, you seem to be trying to use time dilation inappropriately to understand the relativity of simultaneity.
First, it is true that, according to B's clocks, (one in the rear of the train, one in the center, and one in the front), the time it takes for light to travel from the rear to the center and from the front to the center are equal.
In fact, it's true by definition because this how B confirms that his clocks are synchronized; see Einstein synchronization.
But, according to A's (Einstein synchronized) clocks, B's clocks are not synchronized. In fact, according to A, the clock 'in front' is behind the clock 'in back' (though they both run at the same rate).
This is precisely the reason that $\Delta t'_L = \Delta t'_R$ while $\Delta t_L \gt \Delta t_R$; no time dilation is required to explain this - only the constancy of the speed of light and the Einstein synchronization convention.
how is the detailed connection between the statement that the length
measurement has to be simulanous and the quoted derivation?
Sally didn't measure a length, she measured a time and, from that, calculated a length.
Both Sally and Sam agree that their relative speed is $v$ so Sally can calculate the distance between the ends of the platform by recording the elapsed time between the event that she passes one end and the event that she passes the other.
Then, according to her, the length of the platform is $L = v\Delta t_0$.
But this calculated length $L$ is the same she would measure using two synchronized clocks (at rest with respect to Sally) and separated by a distance $L$ (according to a ruler at rest with respect to Sally).
She would find that, according to her, the forward clock is at the end of the platform and the rearward clock is at the beginning of the platform at the same time according to the clocks. For simplicity, stipulate both clocks read 0s when the clocks are at their respective ends of the platform.
Thus, knowing that the end of the platform is distance $L$ away and the speed is $v$, it must be that the end of the platform will pass the rearward clock when the rearward clock reads $\Delta t_0$ seconds.
Length contraction is a direct consequence of time dilation.
I would have to say that both length contraction and time dilation are a consequence of the relativity of simultaneity.
Best Answer
You identified the mistake in your own question. You correctly stated "both time and length should receive a Lorentz transformation", but the mathematical operations that you performed were not a Lorentz transformation. Instead of doing the actual Lorentz transformation you simply multiplied or divided by the Lorentz factor. This is not a Lorentz transform.
The Lorentz transform is derived based on the postulate that the speed of light is invariant. So when used in its correct form you will always get $c$ being invariant. The Lorentz transform is given by:$$t'=\gamma \left(t-\frac{vx}{c^2}\right)$$$$x'=\gamma (x-vt)$$$$y'=y$$$$z'=z$$where$$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
The equation of a pulse of light going out from the origin in the primed frame is $$ c^2 t'^2= x'^2 + y'^2 + z'^2$$ This is a sphere whose radius is $ct'$, meaning that it expands at a speed of $c$. This equation is often called the light cone. By substituting the transformation equations into the light cone equation and simplifying we get $$ c^2 t^2= x^2 + y^2 + z^2$$which shows that the speed of light is the same in all reference frames.