What is the explanation for why comparing fractions by cross-multiplying works?

Really you are multiplying both fractions by the same thing (i.e., $bd$), and then comparing them. One inch is less than two inches, so one mile is less than two miles. Multiplying both quantities by some positive number preserves their relative size.

So $\frac ab<\frac cd$ is, after multiplying both things by $bd$, logically equivalent to $\frac abbd<\frac cdbd$, which simplifies to $ad<bc$. And you can go back again by dividing both things by $bd$.

Let $a,b,c,d$ be positive numbers. To compare the fractions $\frac{a}{b}$ and $\frac{c}{d}$, we first write

$\frac{a}{b} ? \frac{c}{d}$

since we do not yet know whether the correct relation is $=$, $<$, or $>$ we will use "?" for now.

Because $b>0$ and $d>0$, their product $bd$ is also positive. Multiplying both sides by a positive number does not change the inequality sign (this is what basically cross multilication does). Therefore,

$\frac{a}{b}$ ? $\frac{c}{d}$ $\Longleftrightarrow \frac{a}{b}\cdot bd \ ? \frac{c}{d}\cdot bd$

Simplifying both sides gives

$ad$ ? $bc$.

Now $ad$ and $bc$ are positive numbers, so they can be compared directly. If $ad=bc$, then $\frac{a}{b}=\frac{c}{d}$; if $ad<bc$, then $\frac{a}{b}<\frac{c}{d}$; and if $ad>bc$, then $\frac{a}{b}>\frac{c}{d}$.

Thus, cross multiplication works because multiplying both sides of the comparison by the positive number $bd$ converts the comparison of fractions into a comparison of ordinary positive numbers, while preserving the inequality.

In fact, if the fractions are $$x = \frac{a}{b} \quad \text{and} \quad y = \frac{c}{d},$$ then the correct comparison should be $ad$ versus $bc$, not $ab$ versus $cd$.

That said, why does this so-called 'cross multiplication' work? What we are actually doing here is multiplying both fractions by a common denominator, in this case, $bd$. Then we have $$xbd = \frac{a}{b} \cdot bd = ad, \quad \text{and} \quad ybd = \frac{c}{d} \cdot bd = bc.$$ If $b, d$ are positive, then we can compare $x$ and $y$ by equivalently comparing $xbd$ and $ybd$. So if $ad > bc$, then $xbd > ybd$, hence $x > y$; and if $ad < bc$, then $xbd < ybd$, hence $x < y$. And if $ad = bc$, then $x = y$.

When you first work with ratios it's pretty easy to see that $\frac1b \cdot \frac1d = \frac1{bd}$. Why? Well if I cut a pizza into $b$ slices and then each slice into $d$ smaller slices I've made $bd$ slices. It's also easy to see that $\frac11 =\frac{n}n$. All the complications are a consequence of this since $\frac{n}n \cdot \frac{a}b = \frac{na}{nb}$.

So now say we have two fractions and we want to see if they're the same. For that to be the case we'd need them to be of the form $\frac{na}{nb}$ and $\frac{ma}{mb}$ for some $m$ and $n$ with $\frac{a}b$ in lowest terms. If we set those equal we get $\frac{na}{nb}=\frac{ma}{mb}$. Since we may not know what $a,b,n$ or $m$ are in advance division can be difficult. However if we clear denominators we see that we get $nmab=nmab$. That equality is the cross ratio test.

The most fundamental thing to digest is that this relationship defines what equality of rational numbers means for integers a-b-c-d:

$\frac a b = \frac c d$ if and only if $ad = bc$

Keep in mind that historically, and more importantly in the formal mathematical development, there's a point where you have integers but rational numbers (fractions) don't exist yet. This statement of equality is exactly what defines and gives meaning to these new ratio objects, in terms of the previously-known integers: fractions being equal (the new thing) is synonymous with a certain multiplication of two integers (the old thing).

There is no more basic, fundamental meaning to fractions than this. It cannot be devolved or derived from any more basic statement. If you can see why such an axiomatic starting point is necessary, then you have above-average mathematical understanding.

Okay, now, how should I interpret the product of the numerator of the first fraction and the denominator of the second fraction, $ad$ and $bc$? What am I actually doing?

One way to interpret it is that you are bringing the two fractions to equivalent forms having a common denominator:$$\frac{a}{b}=\frac{a}{b}\cdot 1=\frac{a}{b}\cdot\frac{d}{d}=\frac{ad}{bd}$$ $$\frac{c}{d}=\frac{c}{d}\cdot 1=\frac{c}{d}\cdot\frac{b}{b}=\frac{bc}{bd}$$

Therefore, $\frac{a}{b}=\frac{c}{d}$ if and only if $\frac{ad}{bd}$ = $\frac{bc}{bd}$. And because the denominators are the same for the latter pair, the equality holds if and only if the numerators are also equal. That is, if $ad=bc$.

It's slightly more complicated when comparing for inequality ($\gt$, $\lt$), because when $bd$ is negative, canceling it reverses the order of the inequality. The easiest way to approach that is to stipulate that $b$ and $d$ must both be positive. This does not cost us any generality, because if needed, we can always prepare by multiplying one or both of $\frac{a}{b}$ and $\frac{b}{c}$ by $\frac{-1}{-1}=1$ to achieve a form we can work with. Thus, we can say:

Given integers $a$, $b$, $c$, $d$ such that $b,d\gt 0$, the ordering relationship between $\frac{a}{b}$ and $\frac{c}{d}$ is the same as the ordering relationship between $\frac{ad}{bd}$ and $\frac{bc}{bd}$, which, because of the common, positive denominator, is in turn is the same as the relationship between $ad$ and $bc$.