Is Modal, Deontic, and Doxastic Logic Just a Specific Extension of Predicate Logic?

I've done some limited reading into modal, deontic, and doxastic logic. It seems that the for the following qualifiers:

• Modal - $\Box$ (necessarily) and $\Diamond$ (possible);
• Deontic - $\Box$ (must do) and $\Diamond$ (permissible);
• Doxastic - $\mathcal{B}$ (it is believed)

That these can just be written as predicates with given rules, and thus really is a sub-category of predicate logic. Tweaked for aspects of understanding the world.

I've read that modal and deontic seem to operate in the same way as well and thus there is a link there.

Is it true that these can be re-written in forms of predicates with axioms? Are there any greater subtleties to what is going on?

I assume that you are asking about modal (deontic, doxastic, etc.) propositional logic. Your intuition of modal logic being a fragment of predicate logic is right (in which case predicate logic is an extension of modal logic but not the other way around, however), and it can be made precise by using the concept of standard translation. You can see the technical definition on the Wikipedia article, but the point is that, if $M$ is a Kripke model (a relational structure with an interpretation of the propositional variables), then $M , w\Vdash \phi$ if and only if $M \models \mathrm{ST}_x[w]$ where $w \in M$ and $\mathrm{ST}_x(\phi)$ stands for the standard translation of a modal formula $\phi$, which is a first-order formula. Thus, modal logic is a fragment of first-order predicate logic at the level of Kripke models; it is the image of the modal formulas along the function $\mathrm{ST}$. If one is interested in Kripke frames (bare-bone binary structures) on the other hand, we have $F, w \Vdash \phi$ if and only if $F \models \forall \bar P \mathrm{ST}_x(\phi)[w]$, where $\forall \bar P$ stands for a block of second-order quantifiers for the unary predicate symbols occurring in $\mathrm{ST}_x(\phi)$. Thus, modal logic a fragment of second-order predicate logic at the level of Kripke frames.
When it comes to axioms, I can name two subtleties at the level of Kripke frames. First, for the most famous axioms $\phi$ of modal logic, we have a first-order formula $\alpha$ s.t. $F, w \Vdash \phi$ if and only if $F \models \alpha[w]$, despite what I said in the last paragraph. For instance, if $\phi = \Box p \to p$, then we can take $\alpha = Rxx$. In such cases, $\alpha$ is called a first-order correspondent of $\phi$. Of course, not every formula has a first-order correspondent; for instance, $\Box\Diamond p \to \Diamond \Box p$ does not. So if you say "predicates with axioms," you might want to be careful whether you mean first-order predicate logic (which is often the case), or second-order logic.
The second subtlety is about which modal formula follows from a modal axiom. The relation of a modal formula following from another can be made precise in two different ways. In one way, a formula $\phi$ follows from an axiom $\psi$ if and only if for every Kripke frame $F$ we have $F \Vdash \psi \implies F \Vdash \phi$. This is a semantic definition, and it is the standard concept of consequences in second-order logic. In the other way, $\phi$ follows from $\psi$ if and only if $\phi$ is in the least normal modal logic containing $\psi$. This is a syntactic definition. Unfortunately, the two concepts rarely coincide. If they do not, the least normal modal logic containing $\psi$ is called Kripke-incomplete. This is another subtlety you should be careful about.
No, there's an enormous difference between these and predicate logic. The key thing is that in predicate logic, a predicate always applies to an object - a variable, or a constant. In modal, deontic, or doxastic logic, the operators are applied to sentences. Thus $\square Q$ makes sense, but $P(Q)$ doesn't. Objects and sentences are very different things; importantly, the rules of deduction of modal, deontic, and doxastic logic allow deduction inside the operator. So, for example, from $\square P$ and $\square(P \rightarrow Q)$ we can deduce $\square Q$ (under a certain modal axiom system), but there's no situation in predicate logic where from $P(x)$ and $P(y)$ we can deduce $P(z)$.