It is often said, that single clock dilates relatively to a set of spatially separated and Einstein – synchronized clocks.
This set of synchronized clocks is the REST FRAME of the observer. Observer is not a real physical person, but the whole reference frame, or team of observers. Relativistic observer conducts measurements in his own rest frame.
To measure dilation of moving clock an Observer (or team of observers) must have at least two clock in his frame, let's say clock C1 and C2. When moving clock passes by clock C1, he compares readings in immediate vicinity. When clock passes by clock C2, he compares readings again. If moving clock and C1 showed 12 PM at meeting, moving clock will show 3 PM and C2 shows 6 PM when they meet. This is how time dilation works. SINGLE moving clock dilates relatively to a set of synchronized and spatially separated clocks, not vice versa. Set of clock runs faster from the point of view of single clock.
If "moving" observer wants to measure dilation of another clock, he has to turn himself into one "at rest" by means of introducing his own rest frame. He simply puts another additional clock (at least) and synchronizes it by Einstein signalling method (Einstein clock synchronization convention).
https://en.wikipedia.org/wiki/Einstein_synchronisation
This article emphasize this important detail at page 6 (6)
http://isites.harvard.edu/fs/docs/icb.topic455971.files/l09.pdf
Also:
https://arxiv.org/ftp/physics/papers/0512/0512013.pdf
http://www.pstcc.edu/departments/natural_behavioral_sciences/Web%20Physics/Chapter039.htm
"Two spatially separated clocks, A and B, record a greater time interval between two events than the proper time recorded by a single clock that moves from A to B and is present at both events."
It is absolutely clear even visually (animation from article Time Dilation in Wikipedia).
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Notion of Observer in SR https://en.wikipedia.org/wiki/Observer_(special_relativity)
In your case, observer in the reference frame of Earth has to place two Einstein – synchronized clocks. One E1 is at Earth, another E2 is far, far away at the point of arrival of Astronaut. These clocks are Einstein - synchronized. When Astronaut arrives to E2 his clock shows gamma times less time. Obviously, clock E2 shows gamma times more time.
We can imagine a row of synchronized clocks of reference system K. each denoted by letter - A, B, C, D ….. Z. Then a person with single clock on his wrist (A’ for example) moves in this reference system K and compares readings of his clock with these clock A-Z successively. When he comes to the clock Z, his clock A’ shows gamma times less time, than clock Z. Thus, clock Z shows gamma times more time, than his own.
At this point Z this clock A’ immediately turns back and starts travelling in reverse direction, passing by clock Z, Y, X ….. C, B and finally arrives into point A of reference frame K. Clock A’ compares readings with clocks Z-A successively again and sees, that it dilates itself gamma times, i.e. every clock on the way shows gamma times more time. When clock A’ arrives into point A, clock A’ shows gamma times less time than clock A, and clock A shows gamma times more time than clock A’.
This discrepancy of clock readings when they meet again is often called clock paradox or Twin paradox.
All that goes straight from the Lorentz transformations.
$$ T = \frac {t'_{x'}+ \frac {v'} {c^2} x'} {\sqrt {1-( \frac {v} c)^2}} (1)$$
$T$ is clock readings that belongs to reference frame $K$, taken in point $x'$ at moment of time $t'_{x'}$ of reference frame $K'$, and $t'_{x'}$ reading of clocks that belongs to reference frame $K'$ in the point $x'$ of reference frame $K'$
How to interpret Lorentz transform for time?
Transformation demonstrates, that time $T$ of reference frame $K$ (in which it does not depend of $x$ coordinate or any other coordinate) is universal in reference frame $K$ and each point of this frame.
Now let's fix point $x'$, for example $x'=0$. In this case this transformation will look like that:
$$ T= \frac {t'_{0'}} {\sqrt {1-( \frac v c)^2}} (2)$$
$T$ is clock reading of reference frame $K$ taken in point $x'=0$ (in
the origin $O'$ of reference frame $K'$), and $t'_{o'}$ is time in the reference frame $K'$, in particular in the origin $O'$.
We can take $dT/dt'$ when $x'$ is fixed and will get ${dT}/{dt'} = 1/{\sqrt {1- \frac {v^2} {c^2}}} $
According to (2) it is not time $t'_{o'}$ which is showed by single clock in the point $O'$ runs slower, but time $T$ , which is "distributed" through all reference frame $K$ and taken in the origin $O'$ of reference frame $K'$ runs faster (relatively to time $t'_{o'}$ that is in the origin $O'$ of frame $K'$). Time dilation comes by means of transformation of (2) into:
$$t'_{o'}=T \sqrt {1- {\frac {V^2} {c^2}}} $$
It is correct that $T>t'$ and $t'<T$. It is also true that $T'>t$ and $ t<T'$. But that $t<t'$ and $t'<t$ from different points of view is nonsense.
