Twin Paradox: Fewer Seconds or Shorter Seconds? In Einstein's Twin Paradox thought experiment, the travel time is shorter for the traveling brother than for the stationary brother.
Since the SI unit of time in physics is the second, is the travel duration shorter because:

*

*It contains fewer seconds or

*The number of seconds is the same for both brothers, but the traveler's seconds are shorter?

 A: 
It contains fewer seconds

Unambiguously, it contains fewer seconds. The SI Brochure, the defining document produced by the BIPM, explicitly states that the second is a unit of proper time. Specifically, on p. 7 it says

The second, so defined, is the unit of proper time in the sense of the general theory of relativity.

In GR the proper time along some spacetime path $S$ is defined as $$\Delta \tau=\int_S \frac{1}{c} \sqrt{- g_{\mu\nu}dx^\mu dx^\nu}$$ where $g_{\mu\nu}$ is the metric using a $(-+++)$ signature and the $dx$ are the coordinates in which $g$ is expressed.
Evaluating this integral, the traveling twin has fewer seconds, each of which has the same amount of proper time as defined by the SI.
A: John and Dale have given perfectly correct answers, and I am adding mine simply to drive home the point and hopefully to help counter the mis-understandings that arise from the phrase 'moving clocks run slow'. Time dilation is not some effect that impairs the working of clocks so that they under-report time. It is the consequence of the geometry of a 4d spacetime, in which the time interval between two events depends upon the path followed between them.
A: Interesting question. Let's first understand what is actually happening to twins' times.
Twins are A and B. A remains on Earth, B travels with the speed $v$ forward, and then turns at some point and travels back with the speed $-v$.
Let's assume that turn time is negligible compared to the total travel time (i.e. B turns back almost immediately).

Dots on the line $Ob$ correspond to the units of time in the frame of the twin A. Similarly, dots on the line $O\alpha b$ correspond to the units of time in the frame of the twin B.
By counting the dots on both lines, you can see that the line $O\alpha b$ "contains fewer seconds". That is why twin B is actually younger than twin A when they meet at the end of the journey.
To understand why the picture is not symmetric w.r.t. A and B, look at the events $\alpha_1$ and $\alpha_2$. These are the events that are simultaineous (in the reference frame of B) with the event $\alpha$ just before and just after the turn.
It looks like the "simultaneity" jumped from  $\alpha_1$ (which is an earlier event from A's perspective) to $\alpha_2$ (which is much later event from A's perspective).
This effect, however, is not observable.

The second picture shows the light signals sent from A to B (say, A sends to B his photo after every second passed in A's frame). From B's perspective, photos reach him at roughly equal intervals - a bit longer before the turn, and bit shorter after the turn. From B's perspective, A's "seconds" are always "short" during the trip, but they are even "shorter" after the turn then before the turn.
Conclusion: The number of seconds is different. The duration of A's seconds is shorter from B's perspective, and vice versa (symmetrically). All the magic happens during the turn, when "simultaneity" jumpes from  $\alpha_1$ to $\alpha_2$.
A: The two twins start and end at the same point but they trace out different paths in spacetime. Since the paths are different they have different lengths, and the length of the path is equal to the elapsed time shown by a clock held by the observer. There is more on this in the question What is time dilation really? though that might go into too much depth.
So the two path lengths can be measured in seconds, and both twins will agree on the lengths of their path and the other twin's path i.e. they will both agree that the Earth twin's path has more seconds than the moving twin's path. So the answer to your question is:


*

*It contains fewer seconds


A: Less seconds.
Imagine the following...
The two twins stay in radio contact the entire trip. Each twin sends the other twin a message every second (according to the sender's clock). They do this continuously from when the twin in the spaceship leaves till they return.
The count of messages received by the spaceship will be greater than the count of messages sent by the spaceship.
The count of messages received by the twin on earth will be less than the count of messages sent by the twin on earth.
Also...
This effect occurs continuously during the whole trip. It isn't something that occurs all at once at the beginning or the turning point or the end. The messages are steadily sent and steadily received during the whole trip. The rate of receiving them will speed up or slow down, depending on  whether the ship is outbound or inbound, but they will be continuously received.
And on the spaceship there is a panel with two clocks, one called ship time and one called earth time. The ship time clock is a normal clock, while the earth time clock is a clock designed to run faster. Surely these twins know to account for speed and distance as the messages are received, and once accounted for, they will jive with the earth time clock. And the earth time clock on this panel will run as smoothly as the ship time clock, just faster.
My point is that this doesn't happen all at once. It is happening continuously during the entire trip. There are many different ways to analyze these kinds of problems and do the math. If you analyze them all-at-once (at the beginning, middle and end) then you get an all-at-once answer and the impression that the time difference happens all-at-once.
We also know this is a continuous process because we do the exact same thing with our GPS satellites.
https://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=6074806
