Clock $A$ is stationary and Clock $B$ moves around the table. This means that from $A$'s perspective, time will "pass slower" on $B$. It might be tempting to think that the reverse is also true: From $B$'s perspective, $B$ is stationary and $A$ moves around. In principle this is true, i.e. time dilation is symmetric; see Time dilation all messed up! for a nice intuitive understanding using spacetime diagrams.
However, you said that $B$ moves back and forth. In order for this to be possible, $B$'s velocity must change, i.e. $B$ accelerates and decelerates (which both are the same in that they refer to a change of velocity). Acceleration is in a sense absolute, which means that both $A$ and $B$ will agree that it is $B$ that actually accelerates (and not $A$).
This ultimately is the same situation (or rather solution) to the twin paradox. John Rennie has written a very good and detailed explanation at What is the proper way to explain the twin paradox?. The key takeaway is that because $B$ is accelerating, both $A$ and $B$ will agree that time has "passed slower" for $B$. So yes, $B$ will show an earlier time.
Again, this is true. I have far less (if even) experience with general relativity than with special relativity, but the closer one gets to a massive object, the more spacetime is curved and thus time "runs slower". Also, gravitational time dilation is not symmetric so there is no (apparent) "paradox" to be resolved in order to answer whose time will "run slower". Both clocks (or observers, if you wish, since clocks can't think or make measurements) will agree that $B$'s time runs slower, so $B$ will show an earlier time.
Yes, this is true. If we set up a third clock $C$ which is stationary on the table and we will use as a reference, we will find the following:
- $B$'s time "runs slower" compared to $C$ – this is exactly the same as described in situation 2
- $A$'s time "runs slower" compared to $C$ – this is exactly the same as described in situation 1
So yes, it could be possible that, given an appropriate velocity1 of $A$, both $A$ and $B$ will "run slower" by the same amount and will show the same time.
The reason that you most often only hear about special relativistic time dilation for satellites or airplanes is that in these cases, the gravitational time dilation only has a lot less significant effect. Sure, it counteracts special relativistic time dilation to some degree, but not by much.
Yes, this shouldn't make a difference, at least conceptually. The direction of movement of the clock is arbitrary (assuming you wouldn't move it up- or downwards which would bring general relativity into play) – it is only the velocity which determines how much time is dilated.
For moving in circles, it is still the moving clock that accelerates (you somehow have to get it to star moving and eventually stop it). Maybe the magnitude of time dilation will be different when comparing circular to linear motion, but it is in both cases the moving clock whose time "runs slower".
Since this has become a lot longer (though hopefully detailed enough) than originally planned, let me give a short summary for each point:
- Yes, clock $B$'s time will "run slower" as it is moving
- Yes, clock $B$'s time will "run slower" as it is closer to the earth's surface
- Yes, both effects counteract each other and could cancel out – $A$ and $B$ would then show the same time. For "everyday objects" such as satellites, special relativistic time dilation has a lot larger effect than the reduced gravitational time dilation.
- The direction of movement won't affect the fact that the moving clock's time will "run slower".
You may have noticed that I put "time runs slower" and similar expressions in quotes. This is because time is not something that can actually "pass" or "run", so our everyday description of time "passing" is imprecise. See What is time, does it flow, and if so what defines its direction? for more information.
A more formal description of "from $A$'s perspective, $B$'s time runs slower" might be something like "$A$'s measurement of the difference in time between two events2 in the frame of reference where $B$ is stationary will be greater than $B$'s measurements on the same events (if $B$ is moving relative to $A$)". However, this is needlessly long and complicated, so I decided to stick with the more easily understandable formulation.
1 Of course, $A$ accelerates. However, this does still mean that $A$'s time will "run slower" than $C$'s. It only would make computing the time dilation more complicated (compared to $A$ being inertial, i.e. not accelerating).
2 These two events could be, for example, the hands of $B$'s clock being at some positions. From $A$'s perspective, more time passes in $A$'s frame of reference until the second place is reached since $B$'s time "runs slower" from $A$'s perspective. If the two marks were those of two consecutive seconds (i.e. event 1 is "$B$'s second hand is at xx:xx:00" and event two is "$B$'s second hand is at xx:xx:01"), then $B$ will always measure exactly one second between those events. $A$ sees the second hand moving slower and thus they might find that on their clock, 1.2 seconds (this is an arbitrarily chosen number) pass between the two events. This, ultimately, is what time dilation means.