Why "holomorphic" vertex algebra? I have a background quite far from vertex algebras, and it seems like a vertex algebra is holomorphic if basically there is only one irreducible module, namely itself. Why is it called holomorphic?
 A: Those in a hurry can skip to the bottom. Those with some leisure time can read the details.
To explain the answer, it is helpful to remember what vertex algebras are. Suppose you have a quantum field theory. Then among its data is some sort of "algebra" of "point operators" aka "vertex operators", which are the operators that you can apply at points in spacetime. The general "algebra" structure on these operators is slightly subtle, because when you insert operators, their locations matter. The best way to organize the algebra structure is as some sort of factorization algebra, but I won't review that definition. Factorization algebras are basically an updated, more universal version of the Wightman and Haag–Kastler axioms. ​Another way to organize the algebra structure is to record all of the $n$-point functions, and that's the way physicists traditionally do it.
Aside for experts: What you really want is not just a factorization algebra, but (at least) a translation invariant factorization algebra. The translation invariance is extra data — how is the algebra translation invariant — and encodes to the Hamiltonian and momentum operators. Probably you should impose more, like that your algebra be Poincare invariant, which would encode the Lorentz boosts, and perhaps you want to require even the full data of the stress-energy tensor (which tells you how your algebra couples to an infinitesimal background metric).
Example: In quantum mechanics (=0+1D QFT), all operators are point operators, and the algebra thereof is the algebra of all of them is the algebra of operators on your Hilbert space. ("What regularity of operators?" I hear you ask. The answer is that you should package together the various regularities of operators — bounded, unbounded, etc. — into some sort of functional-analytic algebraic object.)
In general, the algebra of point operators does not encode the entire data of the QFT. In general, you would need also to know the extended operators: operators that are inserted along lines, surfaces, etc. Some extended operators are built as integrals aka "descendants" of point operators, but not all of them are; one can contemplate a sort of "index" of the inclusion (point operators and their descendants) $\subset$ (all operators), or more generally of the inclusion (operators of dimension $\leq k$ and their descendants) $\subset$ (all operators). I believe that the following is true, and I know a proof for topological field theories: for an $(n+1)D$ QFT, this inclusion is an iso exactly when the space of vacua is $(n-k-1)$-connected.
In particular, in 1+1D, if there is a unique vacuum, then the algebra of point operators knows the entire QFT.
Now, there is a really fundamental fact about the algebra of all operators in a QFT: its centre is trivial. This fact goes under the name Heisenberg uncertainty principle. What I'm saying is: if you have any operator $a$, and if it is not some multiple of the identity operator, then there is some other operator $b$ whose "commutator" with $a$ is nonzero. I won't try to say what is a "commutator" in general. Rather, I want you to take this as an intuition.
Actually, an even stronger statement holds than just that the centre is trivial. Consider the case of an associative algebra $A$. The centre of $A$ is the endomorphisms of the identity $A$-bimodule $_A A_A$. The stronger statement is that the full category of bimodules is trivial. (Of course, since algebras of operators are functional analytic, you will need to incorporate some analysis into the definition of "bimodule".) This stronger statement implies (and, interpreted correctly, essential is) the Noether theorem: all symmetries of your QFT are inner. (Outline of proof: given a symmetry, write down a bimodule; innerizing the symmetry is the same as trivializing the bimodule.)
In 1+1D, the correct category in place of the "bimodules" are the "vertex modules". In general, any translation-invariant factorization algebra has a notion of "module" supported at a point.
Ok, now let me start to bring this all back to your actual question. Let's take a 1+1D QFT. It has a stress-energy tensor $T_{\mu\nu}$, which is a symmetric 2-tensor. (I mean: it is a vertex operator valued in symmetric 2-tensors.) In 1+1D we can polarize our coordinates into light-cone coordinates: $z = x - t$ and $\bar{z} = x + t$. If you prefer Euclidean signature, Wick-rotate to $y = it$. After polarizing, the stress-energy tensor has the components $T := T_{zz}$, $\bar{T} := T_{\bar{z},\bar{z}}$, and $\operatorname{Tr}(T) := T_{z\bar{z}}$. The field theory is conformal when $\operatorname{Tr}(T) = 0$. In this case, $T$ and $\bar{T}$ commute. Moreover, your full algebra of vertex operators has two distinguished subalgebras: the commutator of $\bar{T}$, and the commutator of $T$.
What is the commutator of $\bar{T}$? Why, it is the subalgebra of those operators $a(z,\bar{z})$ which in fact depend only holomorphically on $z$. In slogans: $[\bar{T}, -] \propto \bar\partial$. In other words, the commutator of $\bar{T}$ is the subalgebra of chiral operators. Call it $V$. Similarly, the commutator of $T$ is the subalebra of antichiral operators. Call it $\bar{W}$.
Then $V$ and $W$ (the latter being the complex conjugate of $\bar{W}$) are both vertex algebras in the sense that you know: they are algebras of operators where the algebra structure depends only holomorphically on the positions of insertions.
So we have our full algebra $A$, and it contains $V \otimes \bar{W}$. By construction, $V$ and $\bar{W}$ commute with each other and are in fact each other's centralizers. This invites you to understand $A$ by decomposing it as a sum of $V \otimes \bar{W}$ modules. Conversely, since $\operatorname{Rep}(A)$ is trivial, and since $V$ and $\bar{W}$ are each other's centralizers, $A$ provides a sort of "Schur–Weyl duality": an identification $F : \operatorname{Rep}(V) \cong \operatorname{Rep}(W)$. Conversely, you can recover $A$ from this identification: indeed, $A = \bigoplus_{M \in \operatorname{Rep}(V)} M \otimes \overline{F(M)}$.
A good working definition of "2D CFT" is "pair of [unitary] vertex operator algebras with an identification between their categories of representations".
Example: When the decomposition of $A$ over $V \otimes \bar{W}$ is finite — heuristically, when $V \otimes \bar{W} \subset A$ is of "finite index" — then your CFT is called rational. When the index is infinite, there remains, as far as I know, some open mathematical questions needed to make this working definition completely rigorous.
Let me explain the reason for the name "rational".
The first CFTs really well understood were the sigma models with target a circle $\mathbb{R}/2\pi r\mathbb{Z}$, or more generally a torus. In this case $V = W$. It turns out that when $r^2$ is irrational (at least I think I want $r^2$ — experts should correct me), then the only chiral operators are the free boson modes. But when $r^2 \in \mathbb{Q}$, there are more chiral operators coming from winding the string around the target, and the index of $V \otimes \bar{V} \subset A$ is basically the height of $r$. The same thing holds more generally for a torus $\mathbb{R}^n/\Lambda$, where what ends up mattering is whether $\lambda^2 \in \mathbb{Q}$ for all $\lambda \in \Lambda$ or not.
Finally, there is a very special class of 2D conformal field theories: the holomorphic conformal field theories. These are the 2D CFTs where not just $\operatorname{Tr}(T) = T_{z\bar{z}}$ but also $\bar{T} = T_{\bar{z}\bar{z}}$ vanishes. Equivalently, these are the CFTs where all operators are chiral aka holomorphic. Note that in this case the algebra I am calling $\bar{W}$ must be trivial, and $V \subset A$ must be an isomorphism. But I told you that $\operatorname{Rep}(A)$ was trivial. So $\operatorname{Rep}(V)$ is trivial.
Aside for experts: Actually, you don't really need $\operatorname{Tr}(T)$ and $\bar{T}$ to vanish. You just need them to be central, which is to say you need them to be multiples of the identity. This is because we really only ever care about their commutators. The values of $\operatorname{Tr}(T)$ and $\bar{T}$, as multiples of the identity, are some sort of "anomalies" or "central charges."
TL;DR: Vertex algebras arise as (sub)algebras of operators in lots of contexts, most importantly in 2D QFT. A very special type of 2D QFT is a holomorphic CFT, which is a 2D QFT where all operators depend only holomorphically (after Wick rotation) on their locations of insertion. Necessary and sufficient conditions for a (unitary) vertex (operator) algebra to arise from a holomorphic CFT is that its category of vertex modules be trivial (i.e. completely reducible and the only irrep is the vacuum module). This is why those vertex algebras are called holomorphic.
