...so I'll try to unconfuse.
Let's draw a simple diagram called a displacement-time diagram. You may have come across these in physics at school - you simply plot the distance of an object from the origin at a given time:
That's the situation you were talking about - the blue line represents the stay-at-home, who stays in the same place as time goes from zero to ten seconds, and the red line represents your traveller who is doing about 0.87 times the speed of light, so travels about 8.7 light seconds in those ten seconds. Additionally, I've marked crosses on each line when their clocks tick - the stay-at-home's clock ticks once per second while (in this frame) the traveller's clock ticks every two seconds.
In high school physics these diagrams are just that - a sometimes useful record of data. But in relativity, we think about
spacetime, which is a four dimensional entity. In this context, these diagrams become something rather more: they are a map of spacetime - or two dimensions of it anyway (and they are usually called Minkowski diagrams). The "time" direction is up the page and the x direction in space is across the page (you can add a second spatial dimension, but we don't need to here because all our motion is in 1d).
Let's add a feature to our diagram:
This version includes horizontal lines, each one of which marks "all of space at one time". For example the line going through the ##t=4\mathrm{s}## mark is the whole of space at the time the (blue) stay-at-home calls four seconds after the start of the experiment. You can see that the (red) traveller is about 3.46 light seconds away at that time.
In pre-relativistic physics that's all there is to it. But Einstein made a key observation: two observers in relative motion
do not agree about what "at one time" means. He also wrote down equations that let us draw lines showing what that the (red) traveller would call "all of space at one time":
And this is the answer to your original question. The traveller's last "now" line, what the traveller calls "all of space 5s after I passed the stay-at-home", passes through the stay-at-home's (blue) line half way between the stay-at-home's 2s and 3s tick. That is, according to the traveller, the stay-at-home's clock ticked 2.5 times "during" five ticks of his own clock, while according to the stay-at-home their clock ticked ten times while the traveller's ticked five times. This is not a paradox: they just mean different things by the word "during", and neither is wrong.
This is also the resolution to the twin paradox, with which you are confusing the above scenario. Let's add a return leg to the above diagram to turn it into the twin paradox:
The stay at home says that "during" the outbound leg the traveller's clock ticked five times; the traveller says that "during" the outbound leg the stay-at-home's clock ticked 2.5 times. And we can draw the "now" lines for the traveller on the return leg too:
Again, you can see that the traveller says that there were 2.5 ticks of the stay-at-home's clock "during" the return leg. But you can also see the error the traveller makes if they simply add those two 2.5s:
they failed to remember that they'd changed direction and therefore changed their definition of "all of space at one time", so they failed to account for the 15 ticks of the stay-at-home's clock between "the end of the outbound leg" and "the beginning of the inbound leg". If they remember that they've changed their definition they can correct for that mistake and will correctly predict that the stay-at-home's clock should read 20s at return.
There are more sophisticated ways of defining "all of space at one time" so that you don't get funny time skips like this (indeed, any approach that yields time skips is basically a mistake). I'm not going to go into them because the maths is a lot messier and this post is long enough already, but there are several ways to do that.
I hope that helps a bit.