In my notes it says, when explaining an immediate consequence of the Lorentz Transformation: The significance of the parameter $v$ is seen as follows. Consider the origin of spatial coordinates $(x', y', z')' = (0, 0, 0)'$ in frame $F'$. From the Lorentz transformation This corresponds to $$ \gamma(v)(x - vt) = 0, y = 0, z = 0 $$ or $\mathbf x = \mathbf v t$, which is the worldline of an object moving uniformly with respect to $F$ with velocity $\mathbf v = (v, 0, 0)$.
Do this mean that they have said where $P$ is w.r.t. $F'$ (origin of $F'$) and they are now trying to find $P$ w.r.t. $F$? It looks as if they are, and if that is the case, are they doing it this way: Do the reverse Lorentz transformation i.e. from frame $F'$ to frame $F$. Or is it: Find out what the worldline of the origin of $F'$ would have to be in frame $F$ in order to give the Lorentz transformation $(x', y', z')' = (0, 0, 0)'$. Does it matter which way we do it? I know it doesn't, but at this stage, it is hard to explain/see why.
Moreover, how can this result be true, when the origin of frame $F'$ is moving with velocity $v$, and the distance between $F$ and $F'$ must therefore be $\gamma(v)$ Also, how comes we are allowed to define such a $v$ existing. $v$, after all, is distance over time, and the "coordinates" or (loosely) distances $|x-x'|$ depend on $v$ itself???
I am guessing that this is a common trap, and I am also guessing that the answer is because we are doing a passive transformation. But that is just a guess, and I'm not sure I even fully understand it...
Also, note that at this stage in my notes, they have only just stated what the definition of the Lorentz transformation is.
As a deeper guess, I am guessing that you can think of the passive transformations as moving from one space-time coordinate to another one, or, a "ghost" in $\mathbb R^3$ Note space-time here still refers to Newton's version.