Sure, I do not know why it is useful but,
1) The timer is started at event (x,t) = (0,0)
The timer is stopped at event (x,t)=(0,-k/(vγ))
1) The timer is started at event (x',t') = (0,0)
The timer is stopped at event (x',t')=(k,-k/v)
I would eliminate gamma since it is a function of v and confuses the issue if we are talking about ①
the v built-in to the problem description or ②
the v built-in to the choice of observer, and say just one of:
$$\begin{pmatrix} x_A \\ t_A \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \begin{pmatrix} x_B \\ t_B \end{pmatrix} = \begin{pmatrix} 0 \\ \frac{-k}{v} \sqrt{1 - \frac{v^2}{c^2}} \end{pmatrix} $$
or
$$\begin{pmatrix} x'_A \\ t'_A \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \begin{pmatrix} x'_B \\ t'_B \end{pmatrix} = \begin{pmatrix} k \\ \frac{-k}{v} \end{pmatrix} $$
and
remember that k/v < 0
and
remember that 0 < |v| < c
and
remember that for both events A and B
$$\begin{pmatrix} x' \\ t' \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \frac{-v}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \frac{-v}{\sqrt{c^2 - v^2}} & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \end{pmatrix} \begin{pmatrix} x \\ t \end{pmatrix}$$.
Remembering all that, we could verify the trivial case for event A (not shown), or the easy case of event B, where
$$ \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \frac{-v}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \frac{-v}{\sqrt{c^2 - v^2}} & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \end{pmatrix} \begin{pmatrix} x_B \\ t_B \end{pmatrix} = \begin{pmatrix} \left( \frac{-v}{\sqrt{1 - \frac{v^2}{c^2}}} \right) \left(\frac{-k}{v} \sqrt{1 - \frac{v^2}{c^2}} \right) \\ \left( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \right) \left( \frac{-k}{v} \sqrt{1 - \frac{v^2}{c^2}} \right) \end{pmatrix} = \begin{pmatrix} k \\ \frac{-k}{v} \end{pmatrix} = \begin{pmatrix} x'_B \\ t'_B \end{pmatrix} $$
But I would not remember what chinglu's interest in the problem was, since it seems self-consistent in problem setup and in evaluation in light of the Lorentz transform.
// Edit -- Oh, yeah. chinglu notices that $$ 0 < t_B - t_A < t'_B - t'_A$$ and claims that this is not time dilation.
But the definition of time dilation is "(relatively) moving clocks tick (relatively) slower". Let's see what this means for this example.
$$t_B - t_A$$ describes the time elapsed on a clock where events A and B happen in the same place relative to a inertial observer, since $$x_A = x_B$$, while $$t'_B - t'_A$$ describes the time elapsed on a clock where the same events A and B happen in
different places.
So if A and B are two events on the worldline of a uniformly moving clock, then $$t_B - t_A$$ describes the
proper time between A and B -- the time that the clock that moves uniformly between A and B measures.
Any clock which measures the same time as the
proper time is at relative rest with the clock that moves uniformly between A and B.
That's the meaning of $$x_A = x_B$$.
Shorter: The observer who sees the clock relatively at rest measures the proper time on his own clock.
But for the other observer, $$x'_A \neq x'_B$$ and therefore sees the clock that moves uniformly between A and B as not at relative rest. For this observer the clock is moving with speed $$ |v| $$. And since $$ 0 < t_B - t_A < t'_B - t'_A$$ it is clear that whatever time this observer sees, it is longer than the
proper time. But the clock that moves uniformly between A and B is the clock that also measures
proper time, and therefore the observer who sees that clock as moving sees the time kept by that moving clock as slower than the time of his own clock.
Shorter: The observer who sees the clock as moving relative to his standard measures a time longer than the proper time on his own clock.
Thus this example does not contradict that moving clocks tick slowly.
Big point: Since relativity is all about relative motion, it is pointless to try and claim one clock is
really at rest because all clocks are at rest with respect to themselves.