Although I have read extensively into both matrhematics and physics, I do not consider myself an abolute authority, so keep that in mind when reading the followiing explanation:
In three dimensional curves, s usually stands for arc length. That is, if a particle goes from a point on a curve that has an s value of say 5 to a point with an s value of 6, the particle has gone one unit length. Using the pythogorian thereom, (s1-s2)^2~= (x1-x2)^2+(y1-y2)^2+(z1-z2)^2 if s is very small. Mathematically, ds represents the concept of the change in s if the change is very small. So |ds|=|s1-s2| and ds=sprt of a bunch of stuff.
Now, if we go to four dimensions, things get more complicated. It’s easy to imagine s in three dimensions to be amount of time that the particle has been moving. After 5 seconds, it has such and such an x coordinate, such and such a y coordinate, etc. But in four dimensions, time becomes a coordinate in which the particle “moves”. That is, the particles “traces” a path through not only height, width, and lenth, but also time.
So what does s represent? I can’t really explain that; the best I can say is that s is the amount time that the particle experiences. So if in a particular frame of reference, a particle spends t seconds moving through d distance (d^2= x^2+y^2+z^2; if your substitute that expression for d^2 in the following equation you’ll end up with the equation of the day), the amount of time that the particle experiences is sprt(t^2-d^2). (Assuming that time and distance are represented in the “same” units).
If you know anything about relativity, this should make at least a modicum of sense to you; the more time a person in one frame of reference experiences, the more everyone experiences. However, the more someone else moves, the faster they must be moving, and therefore the less time they experience. If, in your frame of reference, someone is moving at the speed of light, then d and t will be equal (because the conversion factor between them is defined to be the speed of light) so s^2=0; the person will experience no time.
Another way of looking at this is that space is imaginary time; not imaginary in the sense that it doesn’t exist but imaginary in the mathematical sense of being a multiple of the square root of negative one. Space squared is negative time squared and vice versa.
Another interesting aspect of this equation is that given a particle that is expoeriencing no forces other than gravity, and given two points (in four dimensional space) that you know that the particle visits, you can take all the possible paths between the points and calculate the s value for each path. Whichever path has the greatest s value will be the one that particle takes.
For instance, suppose you have to leave Los Angeles at 8 AM and arrive at New York at exactly 10 PM. Now, in case are not aware, the mass of the earth distorts space-time, making people deep in its gravity well (that is, close to its surface) experience slightly less time that those that are far away from the surface. So if you want to spend the most time in your cross country flight as possible (have some last minute work to get done or something) it might be a good idea to get away from the gravity well. But if you go too far away, and then come back, you’ll have to accelerate to a high velocity to make it out and back in the time alloted. Going at a high speed means that you experience less time. So you want a balance between getting away from the earth’s gravity well and not going to fast. It turns out that the best possible curve is basically a parabola, that is, exactly the path that you would take if you left the earth with a certain velocity, did not accelerate, and just followed a ballistic (that is, unpowered) trajectory to New York. What looks to us like curved paths are actually the longest possible paths, are therefore, in a sense, the “straighest” paths. That is, they are geodesics on the space-time manifolds, to throw out a bunch of fancy terms.
As for the sign difference, that’s just a matter of notation. If you take t^2 to be positive, then that’s “time distance”. A positive time distance between event A and and event B (“event” refers to a point in four dimensional space) represents the most time that it could have taken a signal to go from event A to event B. If the time distance is positive, then in every frame of reference event A will appear to occur before event B. The amount of time will be relative to reference frame, but the order is absolute. If, however, one takes space be positive, then one gets the space distance. If two events have a positive space distance, then it means that there is more space between the two events than can be crossed by any particle moving at a speed equal to or less than light speed. Different reference frames will disagree about which event occurred “first” since the events are “too far apart” to get an absolute order.
I hope this has shed at least some light (no pun intended) onto the issue. If you have specific points you’re still confused about, or clarifications to make, please let me know. I’m sure that there are more elegant ways of explaining this stuff; in fact Feynman probably has written one somewhere.