[OP] The catalytic efficiency of an enzyme is given by $k_\mathrm{cat}/K_\mathrm{M}$ where $k_\mathrm{cat}$ is the turnover number, or the number of molecules that can be produced per second per active site of an enzyme.
The last part is not quite accurate. $k_\mathrm{cat}$ is the rate of the reaction under saturating conditions divided by the enzyme concentration. The dimensions are one divided by time (a first-order rate constant).
[OP] $K_\mathrm{M}$ is a measure of the affinity of the enzyme with the substrate, or the likelihood of binding.
The Michaelis-Menten constant $K_\mathrm{M}$ has a rigorous definition based on rate constants, and its dimensions are the same as those of a concentration. If you interpret $K_\mathrm{M}$ as the affinitiy of the enzyme to the substrate, you have to know that higher values of $K_\mathrm{M}$ correspond to lower degree of binding. The likelihood of binding strongly depends on the concentration of substrate. For discussion of the catalytic efficiency, we are interested in substrate concentrations lower than $K_\mathrm{M}$.
Why bother dividing the $k_\mathrm{cat}$ by $K_\mathrm{M}$? Isn't the affinity of the enzyme already encoded into the quantity of $k_\mathrm{cat}$? How could you be an enzyme that has low affinity, but still have a huge turnover? To me this doesn't seem possible, and thus it is redundant to divide by $K_\mathrm{M}$. Likewise, could there be a situation where $k_\mathrm{cat}$ is low, but $K_\mathrm{M}$ is high?
Here are three examples showing rate vs. substrate concentration. Let's say the red curve shows kinetics of a given enzyme. If we compare the red enzyme to one (in green) that has the same $K_\mathrm{M}$ but a $k_\mathrm{cat}$ smaller by a factor of two, the rate is half the "red" rate at all concentrations. On the other hand, the blue enzyme has the same $k_\mathrm{cat}$ as the red enzyme, but twice the $K_\mathrm{M}$ (remember, this means it is hard to get the substrate to bind). At high substrate concentration, red and blue show the same miaximal rate, but at low concentrations, the "red reaction" is twice as fast as the "blue reaction".
In fact, the green and the blue enzyme show identical behavior at low substrate concentrations because $\frac{k_\mathrm{cat}}{K_\mathrm{M}}$ for the two enzymes is the same. That is the idea of catalytic efficiency.

image source: https://www.desmos.com/calculator/ue4eeg4xyh
Try it yourself
Here is a tool where you can change $K_\mathrm{M}$ (abbreviated as "K" in the tool) and $k_\mathrm{cat}$ (via $v_\mathrm{max}$ abbreviated as "v") to explore how the graph of rate (abbreviated as "y") vs. substrate concentration (abbreviated as "x") changes while varying those two parameters: https://www.desmos.com/calculator/egmxfzxct8
