Why are higher-order logics less well-behaved?

Unfortunately the term is ambiguous: there two kinds of semantics of higher-order languages, and only one is problematic. Consider the language of second-order arithmetic, where there are quantifiers over both natural numbers and sets of natural numbers.

First is what Quine called "set-theory in sheep's clothing": this is where quantification over sets of natural numbers is defined to be over all the sets of number that can be posited. It's the theory we use when we prove that there can be only one complete, totally ordered field. It isn't really a logic, there's no complete notion of proof formalisation for it. Wikipedia calls this "standard semantics"; I'm not sure if there is an authority for this.

Then there's Henkin semantics, which uses the rule analogous to the lambda-calculus to define a semantics for second-order quantifiers. This can be seen as still in the realm of first-order logic, in that sense that the second-order system can be translated into a first-order system conserving provability. This is how second-order arithmetic is defined.

All the theorems of the Henkin semantics will be "theorems" in the first.

Second-order logic does not satisfy the completeness and compactness theorems. Here is a proof that the compactness theorem fails (which itself implies that the completeness theorem fails, because completeness implies compactness). Namely, there is a second-order way (call it $F$) of expressing that a set is finite: every injective function on the set is surjective. This means that if $P$ is a relation on the product set $S \times S$ such that $P$ satisfies the conditions to be a function, and $P(x,y), P(z,y)$ imply $x=z$, then for all $w$ there is $q$ with $P(w,q)$. This statement is quantified over $P$ as well as the variables $x,y,z,w,q$ so is second-order and clearly expresses finiteness.

However, we also have the statement $T_n$ "there exist distinct $x_0, \dots x_n$ in the set," which is even first order, and states that the set has cardinality at least $n$. So the conjunction of $F$ and any finite collection of the $T_n$ is satisfiable, but all of them together cannot be satisfied.