Now to answer why:
We start with one frame of reference, that of the spaceship and set up two events that are one hour apart. I'm assuming that your statement, "An interval of an hour is measured on the spaceship's clock" means that one hour has gone by on the spaceship's clock. This also means that one hour has gone by on all the coordinate clocks in the rest frame of the spaceship, no matter where they are located. So for our two events, we can pick different locations as long as the time interval is one hour. But in the rest frame of the planet, the clocks are running at different rates and so we cannot just pick any two events at different locations, we must make sure that the two events in the spaceship's frame are at the same location in the planet's frame.
Now in picking our two events, the easiest way to do this is to put the first event at the origin and the second event one hour later at some other location. We are going to use the standard configuration of events for the Lorentz Transform.
When doing problems where everything is happening along one line (and we can pretend that the spaceship and planet have no widths so they can pass right through each other), I like to shorten up the nomenclature for an event so that it includes just a time coordinate and an x-coordinate since the y- and z- coordinates are always zero. And I also like to use β for speed, as you did, so that the Lorentz Transform (LT) is much simpler, as follows:
t' = γ(t-βx)
x' = γ(x-βt)
The primed values are for the planet since we are starting with events defined in the spaceship's frame.
You already calculated γ as being equal to 1.25 and you are probably wondering why I said you divide by γ when you can see that the equation says you multiply by γ. Well, the answer to that question is that people like to take short-cuts and in the short-cut, you divide but you'll see when we get done using the LT that it works correctly the way it is written.
OK, so now we define two events in the spacecraft 's frame. The first one will be, as I said, at the origin, with t=0 (in hours) and x=0 (in light-hours). So the coordinates of the first event are:
[0,0]
The second event will be one hour later which we naively might set up as [1,0] which would be an event on the spaceship that was one hour later but in order to do it correctly, as I stated earlier, we need to make sure that we use coordinates in which time is transformed at the same location on the planet, not on the spaceship. So we have to consider where the planet will be one hour later. We know that at time zero, the spacecraft and planet are at the same zero location, so one hour later, the planet will be 0.6 light-hours away since it is traveling at 0.6c, so this makes the event in the spaceship's frame of where the spaceship will be one hour later as:
[1,0.6]
Now we need to transform both these events from the spaceship's frame to the planet's frame but since we already know that the origins coincide (in other words, [0,0] transforms into [0,0] in any other frame), we don't have to do any work to transform the first event.
Now for the second event, [1,0.6], we have the following values:
t=1
x=0.6
β=0.6
γ=1.25
Solving for t' we have;
t' = γ(t-βx)
t' = 1.25(1-0.6*0.6)
t' = 1.25(1-0.36)
t' = 1.25(.64)
t' = 0.8
And 0.8 light-hours is 48 minutes which is what you get when you divide 1 hour by 1.25.
OK, now let's do x':
x' = γ(x-βt)
x' = 1.25(0.6-0.6*1)
x' = 1.25(0.6-0.6)
x' = 1.25(0)
x' = 0
Technically, when working with time intervals, we should subtract the two times that we calculated that define the starting and ending times of the interval, but since the starting time was at the origin (t=0) we don't have to explicitly do the subtraction.
Now to summarize, in the spaceship's frame, we have two events separated by one hour:
[0,0]
[1,0.6]
These transform into the planet's frame as:
[0,0]
[0.8,0]
We can see that in the planet's frame, the two events are at the same location, x=0, and we can see that the time has advanced by 0.8 hours.
