Question 1: Interpretation
Ok, so here is how I like to think of it from a geometric perspective.
Definition: (Under moderate hypotheses) analytic spread of an ideal $J$ is the minimum number of generators of an ideal with the same integral closure as $J$.
Definition: Two ideals $J, J'$ have same integral integral closure if and only if their normalized blowups $X = X'$ are the same with equal pullbacks $J \cdot O_X = J' \cdot O_X$.
The number of generators of an ideal of course gives a bound on the number of affine charts you need to cover a blowup.
Putting these together you obtain that:
Fact: Analytic spread of $J$ is a canonical upper bound on the number of affine charts you need to cover the blowup of $J$. In other words, it measures the complexity of a blowup of $J$.
Let's look at the usual:
Example: Consider $J = \langle x^3, x^2y, xy^2, y^3 \rangle \subseteq k[x,y]$. Then the analytic spread is two, corresponding to the ideal $J' = \langle x^3, y^3 \rangle$. The normalized blowup of $J'$ is the same as the blowup of $J$, which is just blowing up the origin in $\mathbb{A}^2$. On the other hand, you only need two charts to cover the blowup of $J$, the chart corresponding to inverting $x^3$, and the chart corresponding to inverting $y^3$.
Now, it could be that the analytic spread is exactly the minimum number of charts needed to cover the blowup, I don't know (if it's important I could think of a couple people I would ask).
Finally, if you were looking for more algebraic interpretations, I would see if anything in the Swanson-Huneke book on integral closure helps.
Question 2: Computing it
Did you try:
http://www.math.uiuc.edu/Macaulay2/doc/Macaulay2-1.6/share/doc/Macaulay2/ReesAlgebra/html/_analytic__Spread.html