Suppose we have a velocity vector field given by $\mathbf v=k_1x^2\ \mathbf{i}+k_2y\ \mathbf{j}+k_3z\ \mathbf{k}$.
$\displaystyle\text{div }{v}=\frac{\partial(k_1x^2)}{\partial x}+\frac{\partial(k_2y)}{\partial y}+\frac{\partial(k_3z)}{\partial z}=2k_1x+k_2+k_3$
At $(0, 0, 0)$, $\text{div }{v}=k_2+k_3$. Now $k_2+k_3$ has dimensions of $[T^{-1}]$.
What is the physical interpretation of the value of $\text{div }{v}$ at the point $(0, 0, 0)$?
If you add a constant vector to $v$, the divergence doesn't change. So it's nice to look at the case where $v(x_0, y_0, z_0) = (0,0,0)$.
In this case, it roughly measures the following:
Put a small box (or sphere) $R$ (for "region") around your point $P = (x_0, y_0, z_0)$. Let every point within $R$ "flow" along the vector field for some small time $t$, to get a new region $R'$. Compare the volume of $R'$ to that of $R$, as a function of time and look at a Taylor series: $$ \frac{vol(R'(t))}{vol(R(t))} = 1 + kt + m \frac{t^2}{2!} + \ldots $$
The number $k$ is (roughly) the divergence of $v$: it's the instantaneous rate of change of volume near $P$. I say "roughly" because you have to take a limit of the $k$-values you get, as you shrink the region $R$ to be smaller and smaller.
There are several subtleties hidden here. First, you need to know that $v$ is smooth enough that the region $R'$ actually has a volume, rather than becoming some weird shape that's not measurable. But that's not too tough.
Second, you need to know that the volume of $R'$, as a function of $t$, is smooth enough to have a Taylor series, at least a first-order Taylor series. That, too, isn't too tough: it follows from $v$ being differentiable.
Third, you might wonder why the ratio of volumes doesn't care whether we use a sphere or a cube or a dodecahedron, etc. It actually does care: it's critical that the region $R$ have nonzero volume, or everything falls apart. So $R$ cannot be just a square in the $xy$-plane, for instance. Once you know $R$ has nonzero volume, you know it contains some ball, which contains some cube, which contains another ball, and so on, so the cube/ball difference turns out not to matter.
Fourth, I suspect that it's important that you pick $R$ to not be "too big": all of $R$ should fit within some ball. That makes the "shrinking" process certain to have a limit.
But the intuition is still what I said: it's the instantaneous change of volume near $P$.
What if $v(P)$ is not $(0,0,0)$? Then you can imagine fluid flowing according to $v$, and if you "ride" on a particle that's at position $P$ at time $t = 0$, and observe how the fluid volume around you is expanding or contracting at time $0$, you'll have found the divergence (or something very close to it -- the so-called "material derivative"'s second term may have an impact here).
I once saw all this explained by a physicist (perhaps Aaron Lemonick?) who said "imagine that $v$ represents the flow of mashed potatoes. You surround a bunch of potatoes with a pair of colanders that mate up to form a sphere centered at $P$, and then as the potatoes flow, carrying along the colanders, either some mashed potatoes squeeze out the holes in the colander, or get sucked in (or both, in different places). The divergence measures the net flow (at least if the colanders are very very small)."