Can cobordisms of 3 or 4 manifolds be visualized by moves on kirby diagrams? I'm mostly interested in the 4d case so I'll state the question in that form.  Basically, it comes down to two parts:
1) What simple moves can be performed to a kirby diagram of a 4-manifold that guarantee that the resulting diagram is cobordant (via a 5d manifold) to the original manifold.
2) Can any cobordism be achieved through such moves?
 A: I think that a complete set of moves for cobordisms are the following: Kirby moves (that preserve the 4-manifold), handle trading (dotted circles become 0-framed 2-handles), addition/deletion of pairs of trivial circles with framings $\pm 1$ (corresponding to connected sums with $CP^2 \,\#\, {\overline{CP^2}}$). This could be proved by using Kirby calculus for 3-manifolds (which is exact!) as in Kirby's original paper as follows. Take two cobordant 4-manifolds $M_1$ and $M_2$ with their Kirby diagrams (with one 0-handle and one 4-handle). Up to addition of complementary 2/3-handles, we can assume that they have the same number of 3-handles. Now, removing the 3- and 4-handles, you get 4D 2-handlebodies with the same boundary $\#_k\, (S^1 \times S^2)$ for some $k \geq 0$. Trading 1-handles completely remove them (without changing the boundary and the signature, because it is surgery along embedded circles). By Kirby's theorem, the resulting manifolds can be related by Kirby moves (now only on 2-handles) and addition/deletion of separated $\pm 1$-framed trivial knots. When the latter operation occurs, just compensate by adding another suitable $\mp 1$ trivial circle that will be not involved in subsequent moves. You will end up with the Kirby diagram for $M_2$ (maybe after 1-handle trading) plus some $\pm1$ trivial knots (the ones you added for compensation) and the $\pm 1$'s sum up to 0 since the signature must be the same.
A: I'm not sure what you mean by moves, but you can say something. As Marco says, a 5-dimensional cobordism can be decomposed into a series of handle additions, so you have to see what each handle addition does. Each of those is a surgery on an embedded sphere. Let's pretend there are no 0 and 5-handles, these are kind of trivial and what follows can be easily modified. Much of this is in Gompf-Stipsicz [GS] or in Akbulut's forthcoming book; I assume you are already familiar with the basics.
A 1-handle addition is given by connected sum with $S^1 \times S^3$ so you'd add a 4d 1-handle to your diagram. There should be a 4d 3-handle as well but we don't bother. A 5D 2-handle results in surgery on your 4-manifold along a curve C. To see the diagrammatic effect, do 4d handle moves until C becomes the core of a 1-handle; surgery then changes that dotted 1-handle to a 2-handle. You have to think a little about the framing of the surgery to determine the framing of that 2-handle.
It starts to get tricky when you add a 5d 3-handle, since you have to locate a 2-sphere. I don't know an algorithmic procedure, but I think you should be able to find some examples of describing 2-spheres eg in Gompf's paper on the Akbulut-Kirby sphere, Topology {\bf 30} 1991, 97-115. Once you find the sphere, you're not out of the woods; the best method I could think of is to use standard methods [GS] to expand your given diagram to a handle diagram of the complement, and then add in a 3-handle to carry out the replacement of $S^2 \times D^2$ with $S^1 \times B^3$. Doesn't sound like much fun to me!
Finally, for 5d 4-handles, you'd have to locate a 3-sphere as an attaching region. This is also not something very algorithmic; you'd probably have to do 4d handle slides to reveal your manifold as a connected sum or to find a non-separating 3-sphere. 
