What's the difference between the rank and the degree of a type function?
The book states that the rank of a term is the number of $\Pi$'s in its (normal) form. But doesn't the number of such $\Pi$'s correspond exactly to the number of arguments it takes? That is, if a type function takes $n$ arguments, then doesn't that just mean there must be $n$ $\Pi$ symbols to capture those $n$ arguments?


Nope. Let $\sigma, \tau$ be types (i.e., type functions with degree=rank=$0$). The type
$$
\prod_{x:\sigma} \tau
$$
is the type of functions $\sigma \to \tau$, which do not take type arguments -- they take value arguments (of type $\sigma$). So, the degree is $0$ (not $1$), while the rank is $1$.