We have one physical system, the clock, and two coordinate systems, Oc and Os. As I read your question Oc and the inside observer there moves with speed v relative to Os, which by default designates that coordinate system as static. We'll call the observer in the Os coordinate system the outside observer.
The inside observer sees the clock as stationary as the system and the observer are in the same Oc. So the insider continues to see dt = 4 seconds by his Rolex.
Now let's see what the outsider sees using his magic telescope. For "simplicity" since we're into that, we'll assume the outsider has a broadside point of view much like the view in the image. So the clock is whizzing by the outsider in the direction AC.
So the insider sees a projection of the photon onto BD as it leaves and then returns to A. The period of that projection is dt = 4 seconds according to the insider's Rollex. The system and the observer are in the same coordinate system Oc. And the projection's pathway ds is up and down over that BD leg.
The outsider with his magic telescope sees the same period as dT by his Timex.
In that time dT, the outsider also sees the clock move X = v dT relative to the Os coordinate system in the AC direction. But that also results in the outsider seeing the projected photon move a path relative to Os as c dT.
So what we have is a right triangle of distances the projected photon traveled when the clock advances X in time dT. The projected photon along BD and back, travels ds = c dt over one period. And the clock goes X = v dT in the same period.
X and ds are the two sides of that right triangle. The apparent motion of the projected photon due to the clock's motion, dS = c dT, is the hypotenuse. And here's the issue.
That projected photon is traveling at the same speed, c, no matter who observes it. It's c when the insider clocks it and it's c when the outsider does it at well. So we have dS = c dT and ds = c dt as the distance traveled by the projected photon when seen by the outside and as the distance traveled when seen by the insider, respectively. But its the same clock for both the insider and the outsider. So dS = ds = 2BD should be the case.
But it isn't.
dS is the hypotenuse of the right triangle. And that's always greater than either of the two sides. So we have dS = c dT > c dt = ds or, ta da, dT > dt. The rate of time aboard the clock coordinate system when clocked relative to the static coordinate system is slower in the moving coordinate system Oc.
In fact were we to invoke the Pythagorean Theorem for the three sides of our observed right triangle and rearrange terms, we'd find dT = dt/sqrt(1 - (v/c)^2) which is the usual time dilation equation.
Sorry it's not an anti dilation clock at all.