Everything is clear once you remember that simultaneity is relative.

Let's look at a simple case where frame F' is moving to the right with velocity v with respect to frame F, in another words, F is moving at -v with respect to F'.

The Lorentz transformation in one direction is given by \begin{align} x' &= \gamma (x-vt) \\ t' &= \gamma (t - \frac{v}{c^2}x) \end{align}

If you fix $x'=0$, then $t'$ would be the time of the clock inside F'. Simple substitution shows \begin{align} x=vt &\Rightarrow t'=\gamma(t-\frac{v^2}{c^2}t)\\ &\Rightarrow t'=\frac{t}{\gamma} \end{align}

We usually denote this $t'$ with $\tau$ and call it the proper time because the "clock" is fixed to $x'=0$ in frame F'. We can see the time dilation in this case.

However, at time $\tau$ in frame F', we can ask the question of what does the observer at frame F' see at this moment on the clock sitting at the origin frame F? The key thing here is that at this moment means different things to F' and to F due to the relativity of simultaneity. In the Minkowski diagram, it is tilted.

If we fix $t'=\tau$ for some value of $\tau$, we can figure out what $t$ is at $x=0$. \begin{align} \tau&=\gamma(t-0)\\ \frac{\tau}{\gamma}&= t \end{align} Here, we can denote $t$ with something else such as $\overline{t}$ meaning it is what the F' observer sees on the clock at the origin of F. You can see that both observers think that there is a slowdown and at the root it works because simultaneity is relative.