I have trouble understanding the Lorentz transformation to proof the
dilation of time.
Let's use finite differences instead and, further, the entire expression for $\Delta t'$ from the Lorentz transformation
$$\Delta t' = \gamma \left(\Delta t - \frac{v}{c^2}\Delta x \right) = \frac{\Delta t - \frac{v}{c^2}\Delta x}{\sqrt{1 - \frac{v^2}{c^2}}}$$
where $v$ is the relative speed of the primed and unprimed systems.
Now, assume $\Delta x = 0$ which means that the two events are co-located in the unprimed system. So, for example, this would be the case for a clock at rest in the unprimed system. It follows that $\Delta t$ is, in this case, the elapsed time according to a clock at rest in the unprimed system.
Now, this clock at rest in the unprimed system has speed $v$ in the primed system thus,
- $\Delta t$ is the elapsed time according to a clock moving with speed
$v$ in the primed system.
Then, according to the equation above
$$\Delta t' = \frac{\Delta t - \frac{v}{c^2}\cdot 0}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}}$$
$\Delta t$ is smaller than the elapsed time according to clocks at rest in the primed system.
Once again, $\Delta t$ is the elapsed time according to a clock moving with speed $v$ in the primed system and, according to clocks at rest in the primed system, this elapsed time is less than the elapsed time in the primed system.
In other words, moving clocks run slower than clocks at rest. This is time dilation (due to uniform relative motion).