Both Frames Moving Linearly...

And if both reference frames are moving,

From the post "Thought Experiment Gone Wrong" dated 17 Nov 2017,

$x^{'}=\cfrac{x-vt}{\sqrt{1-v^2}}$

So,

$x^{''}=\cfrac{x^{'}-v^{'}t}{\sqrt{1-{v^{'}}^{2}}}$

It is tempting to substitute one into the other, and get

$x^{"}=\cfrac{\cfrac{x-vt}{\sqrt{1-v^2}}-v^{'}t}{\sqrt{1-{v^{'}}^{2}}}$

in this case, $v^{'}$ is the speed measured in $x^{'}$ frame not $x$.  Every unit in $x^{'}$ as measured in $x$ is scaled by the factor $\cfrac{1}{\sqrt{1-v^2}}$.  Instead we should have,

$x^{"}=\cfrac{{x-vt}-v^{'}t}{\sqrt{({1-v^2})(1-{v^{'}}^{2})}}=\cfrac{x-(v+v^{'})t}{\sqrt{({1-v^2})(1-{v^{'}}^{2})}}$

where $v^{'}$ is also scaled by $\cfrac{1}{\sqrt{1-v^2}}$,

$v^{'}\rightarrow \cfrac{v^{'}}{\sqrt{1-v^2}}$.

Do we scale $v^{'}$ again in the denominator?  No.  The scaling factor from $x^{"}$ to $x^{'}$ does not get scaled again.  The translation term get scaled.

Just for the fun of it...