Just as $k_{cat}$ represents the rate of reaction at saturating substrate concentration, $k_{cat} / K_m$ represents the rate of the reaction at negligible substrate concentration.
If we take a look at the standard one substrate/one product Michaelis–Menten kinetics rate equation:
$$v = \frac{k_{cat}[E][S]}{K_m + [S]}$$
We can imagine what happens when $[S] \to 0$, we see that when $[S] \ll K_m$, the denomiator can be reduced to $K_m$, and thus the rate equation becomes
$$v = \frac{k_{cat}}{K_m}[E][S]$$
Or in other words, $k_{cat} / K_m$ is the (pseudo-)second order rate constant between the enzyme and the substrate, when $[S] \ll K_m$.
In fact, when you get into more complex rate equations (like inhibitors, pH effects, and kinetic isotope effects) there's a decent argument (one often made by W.W. Cleland) to be made that the two key constants for enzymatic reactions are $k_{cat}$ and $k_{cat} / K_m$, and it's $K_m$ that should be thought of as the "derived" constant. (The fact that we write the rate equation in terms of $k_{cat}$ and $K_m$ is a historical accident - we could have just as easily had $ v = \frac{k_{cat}\Phi [E][S]}{K_m + \Phi [S]}$, where I've arbitrarily chosen $\Phi$ as the symbol for $k_{cat}/K_m$.)
This still leaves the issue of why $k_{cat} / K_m$ is often referred to as the "specificity constant" of the enzyme. The reason for this is that if you have a single enzyme in the presence of two different substrates, you have a competitive inhibition setup. The math is a little dense (see here or here for examples), but the end result is that the ratio of reaction rates for the two substrates is related to the ratio of their respective $k_{cat} / K_m$'s:
$$\frac{v_a}{v_b} = \frac{k_{cat,A} / K_{m,A}}{k_{cat,B} / K_{m,B}}\frac{[A]}{[B]}$$
This actually makes intuitive sense, with the right mindset - the only stage where the two substrates are competing (where you make the decision to do reaction A versus reaction B) is when they're binding to free enzyme. "The rate when you only care about free enzyme" is equivalent to the negligible substrate case. With negligible substrate, all you have is free enzyme - there isn't enough substrate to have appreciable amounts of substrate-bound form, and the rate of enzyme-substrate encounter is much lower than the rate of product formation, meaning that in the steady state you don't have appreciable amounts of product-bound form, either. Thus the $k_{cat} / K_m$ for a particular substrate is representing how good the free enzyme is at performing that reaction.
