Lorentz transformation of a frequency modulated signal Let's consider the following problem:
A spacecraft starts at time $t_0 = 0$ with speed $v > 0$ and moves along the $x$-axis. If the distance between spacecraft and earth equals $R$ (in the reference frame of the earth), i.e. at $t=\frac{R}{v}$ a frequency modulatated signal
$$
u \colon \left[ \frac{R}{v}, \frac{R}{v} + \tau \right] \rightarrow \mathbb{R} \\
u(t) \colon= u_0 \cos\left( F\left(t - \frac{R}{v}\right) \right)
$$
of duration $\tau$ is sent from earth to the spacecraft. To compute the signal received from the spacecraft we have to consider the spacetime-signal
$$
u(t, x) = u_0 \cos\left( F\left(t - \frac{R}{v} - \frac{x}{c}\right) \right)
$$
and use the Lorentz transformation, i.e.,
$$
t^\prime = \gamma \left( t + \frac{vx}{c^2} \right)\\
x^\prime = \gamma \left( x + vt \right).
$$
Hence, we get the transformation
$$
\left(t - \frac{x}{c}\right) \longrightarrow \sqrt{\frac{c-v}{c+v}} \left(t - \frac{x}{c}\right),
$$
and clearly see the Doppler shift.
Let's come to the actual question:
Assume that the spacecraft did not start at $t_0=0$ from earth. Instead an observer from earth knows that the distance to the spacecraft at $t_0=0$ is $R$ and that it moves with $v\geqslant 0$.
Question 1: What is the correct Lorentz transform in that case? For example we could just shift the time, i.e.,
$$
t^\prime = \gamma \left( t + \frac{R}{v} + \frac{vx}{c^2} \right)\\
x^\prime = \gamma \left( x + R + vt \right),
$$
but obviously the limiting case $v \to 0$ does not exist.
Question 2: A spatial shift instead of a time shift, i.e.,
$$
t^\prime = \gamma \left( t + \frac{v(x+R)}{c^2} \right)\\
x^\prime = \gamma \left( x + R + vt \right),
$$
is somewhat confusing for me. At least the limiting case $v \to 0$ exists in this case, but what is the inverse of this transformation if the spacecraft also knows that the distance to earth at $t_0 = 0$ is $R$ (in the reference frame of the earth). Probably I have a wrong understanding about simultaneity and synchronization of both reference frames...
Maybe someone can shed light on the darkness :)
 A: 
What is the correct Lorentz transform in that case?

There is no need to change the Lorentz transform at all in that case. It is perfectly fine for a ship to be located somewhere other than the origin. In fact, even in the original scenario although the ship starts at the origin by the time it finishes the signal it is no longer at the origin. That is not important in either scenario.
However, if you want to transform it then there is no problem doing a spatial and/or temporal translation. The order of operations matters, it is different to do a translation first followed by a boost vs a boost followed by a translation. Based on the description, I think that a translation followed by a boost makes more sense. The overall combined transform would be: $$ t'= \gamma \left( t+ \Delta t +\frac{v \ (x+\Delta x)}{c^2} \right)$$ $$x' = \gamma \left( x+ \Delta x + v \ (t+\Delta t) \right)$$
For your scenario as I understand it $\Delta x = R$ and $\Delta t = 0$ so the above simplifies to $$ t'= \gamma \left( t +\frac{v \ (x+R)}{c^2} \right)$$ $$x' = \gamma \left( x+ R + v \ t \right)$$ Note that this agrees with your second expression.
Now, your first expression is a little odd. What you are doing is shifting in time, but you are shifting in time by the time that it takes for the ship to arrive at the origin traveling a distance of $R$ at a speed $v$. If $v=0$ then there is no such time. This has nothing to do with relativity, the same thing would happen in Newtonian physics. If $v=0$ then the ship stays at $R$ and never arrives at Earth.

what is the inverse of this transformation

The inverse transform is found simply by algebraically solving the forward transform listed above: $$t=\gamma \left( t' - \frac{v \ x'}{c^2} \right) - \Delta t$$ $$x = \gamma \left( x'- v \ t' \right) -\Delta x $$ Note that this is not simply the first transformation listed with the signs reversed.
That is because the order of transformations matters. A translation followed by a boost is different from a boost followed by a translation. And the inverse of a translation followed by a boost is not a translation followed by a boost. In fact, the inverse of a translation followed by a boost is a boost followed by a translation. That is the form that we see in the final expression.
