"Moving" clocks In regard to relativity of simultaneity, why are clocks in the "moving" frame of reference unsynchronized (i.e., in the direction of motion, they run behind toward the front and ahead toward the rear)? The math explains it, but does not allow me to "see" it intuitively. Can anyone explain the precise reason or reasons in non mathematical terms. For example, do length contraction and time dilation have any role in producing this phenomenon?
 A: Here's a position-vs-time graph (time running upwards) [a.k.a. a spacetime diagram]
that can help you visualize what is going on.
The "rotated graph paper" helps us visualize the tickmarks on the diagram.
The grid of “light-clock diamonds” is based on the trace of the light-signals in Alice's longitudinal light-clock (not the typical textbook's transverse light-clock).
Alice is an inertial astronaut who is at rest in the diagram.
The grid for another inertial astronaut has light-clock diamonds with edges parallel with this grid (the lightlike directions are eigenvectors of the boost) and has equal area (since the determinant of the boost (the product of the eigenvalues) equals 1). The timelike diagonal of the diamond is along the worldline of the other inertial astronaut; the spacelike diagonal is simultaneous according to that astronaut.

Inertial astronaut Bob travels with velocity (3/5)c according to Alice. (So, $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}=5/4$ and $k=\sqrt{\frac{1+(v/c)}{1-(v/c)}}=2$).
The front and back of Bob's train are shown. 
The proper length of Bob's train (the distance between the worldlines according to Bob [along BF]) is 10 space-ticks (sticks) long. It is 8 ticks according to Alice... check by counting diamonds.
According to Bob, events B and F are simultaneous (and simultaneous with event O) and are both 5 sticks away from O.

*

*Consider the following radar experiment.
Bob sent a light signal when his clock read $t_S=-5\mbox{ ticks}$
and received an echo from both the front and rear ends when his clock read $t_R=5\mbox{ ticks}$.

The reflections must have occurred at the "mid-time" $t_B=t_F=\frac{1}{2}(t_R+t_S)=0\mbox{ ticks}=t_O$ and at locations $x_F=\frac{1}{2}c(t_R-t_S)=5\mbox{ sticks}$ (half the round-trip time multiplied by $c$) and similarly $x_B=-5\mbox{ sticks}$. (B and F are in the intersection of the future light cone of S and the past light cone of R.

The intersection of these light-cones-with-their-interiors is called the “causal diamond of S and R”. These causal diamonds are scaled versions of the “light-clock diamonds” along the inertial segment from S to R.)

However, according to Alice, the [spacelike-separated] events occurred in the order B, then O, then F, when Alice's clock read -4, 0,and 4, respectively. Check by counting.

*

*This implies that: when Bob (at the center of the train) met Alice at event O, according to Alice, the rear clock is "ahead" (it reads 3 ticks) and the front clock is "behind" (it reads -3 ticks).

A: I'd take it from the invariant interval (in the $i^{th}$) frame:
$$ \Delta s^2 = (c\Delta t_i)^2 - (\Delta x_i)^2 $$
doesn't depend on $i$.
So in a frame where $\Delta t_i=0$, that is, two distant events are simultaneous, any moving frame will see them at different times.
A: It's better to consider time dilation and length contraction to be the results of the relativity of simultaneity rather than the cause of it. You might find it useful to consider purely spatial analogies. Imagine you live in New York and I live in London. Your z axis is a line perpendicular to the surface of the Earth where you are, and points in a different direction to my z axis in London. What you might consider to be two level points in your frame of reference will be points at two different heights in my frame of reference, owing to the fact that your frame of reference is tilted relative to mine. The relativity of simultaneity is just the same sort of idea extended to four dimensional spacetime. If you and I are moving relative to each other, your time axis is tilted relative to mine, so a horizontal slice through time in your frame is a sloping slice through time in mine.
A: Ultimately, the synchronization of the clocks comes from sending a signal from one to the other. You send a light signal from one clock to another, and you set the time of the clock that received the signal to be equal to the time of the sending clock plus (distance between them)/c. If you use a signal that's travelling at a speed other than c, the math is more complicated, but it ends up at the same place.
If Alice and Bob are moving at velocity v with respect to each other, and Alice sets up a bunch of clocks that are, in her frame of reference, synchronous and stationary, then in Bob's frame of reference, the clocks are not stationary. By the time the signal goes from the sending clock to the receiving clock, the receiving clock isn't in the same place it was when the sending clock sent out the light. So Bob disagrees with the distance between them, and so disagrees with what the correct (distance between them)/c offset is.
A: 
You can visualise two light clocks in front of you then imagine moving one to the right.  The light now takes a longer path and so completing one cycle (while it was moving) takes longer (than the one standing still) and the clocks are now out of sync.
In summary moving a light clock effects the length of the path light takes to complete one cycle and so effects the time it keeps.
You can also picture a stationary observer looking at a moving train with two light clocks in the center.  The stationary observer will see the light taking a diagonal up-down path.  Moving a clock to the back of the train will compress this path (from their perspective) and moving it to the front will expand the path (while the clock is moving).  This shows intuitively how the clocks get out of sync between front and back of a moving train.
It is important to note that there is nothing special about a light clock and that all clocks will be similarly effected (and in fact any physical process).
