My PhD work used C*-algebras quite heavily, so I guess I can claim some expertise there, but I'm not an expert in QFT. That will be the main perspective of my answer.

A good starting point for this discussion is the Stone-von Neumann theorem, a foundational result in both operator algebras and quantum mechanics. The setup is basically the Heisenberg uncertainty principle, which asserts that the operations of measuring the position $x$ and the momentum $p$ of a quantum system don't commute:

$$[x,p] = 2\pi i h$$

An important *mathematical* question about quantum mechanics in its early history was: what kind of objects are $x$ and $p$? Physicists want them to be self-adjoint operators on some Hilbert space, but you can prove rigorously that no pair of bounded operators have this property. This result belongs to the representation theory of Lie algebras - essentially, the Lie algebra with two generators and the relation above has no representation by bounded self-adjoint operators on Hilbert space.

Stone and von Neumann's idea was to focus on the Lie group rather than the Lie algebra; the relation above is the derivative at 0 of the following relation between time evolution operators $U(t)$ and $V(s)$:

$$U(t) V(s) = e^{-ist} V(s) U(t)$$

The Lie group generated by such $U$ and $V$ is called the *Heisenberg group*, and the Stone-von-Neumann theorem asserts that that this group has a unique unitary representation on Hilbert space, up to unitary equivalence (and some adjectives that I won't go into here). This provides a nice foundation for basic quantum mechanics which unifies the Heisenberg and Schrodinger pictures of the theory into one set of axioms.

To handle more complicated quantum systems, we need to generalize to more operators satisfying possibly more complicated relations. Here's how this generalization works:

- Start with a locally compact group $G$; for the original Stone-von-Neumann theorem, $G = \mathbb{R}$.
- The Fourier transform determines and isomorphism $C^*(G) \to C_0(\hat{G})$, where $C^*(G)$ is the group C*-algebra and $\hat{G}$ is the Pontryagin dual.
- Such an isomorphism is equivalent to a unitary representation of the crossed-product algebra $C_0(G) \rtimes G$.
- All irreps of this C*-algebra are unitarily equivalent.

So now we have quantum mechanics for systems with many particles. But what about QFT? The basic reason why QFT is hard, as I understand it, is that the Stone-von-Neumann theorem is no longer true.

For ordinary quantum mechanics, the classical phase spaces are finite dimensional manifolds - for instance, the classical phase space of a single particle flying around in $\mathbb{R}^3$ is $\mathbb{R}^6$. The classical analog of the phase space in quantum field theory, however, is the space of paths in $\mathbb{R}^3$, which is some sort of infinite dimensional manifold. This means infinitely many operators with infinitely many commutation relations, and the corresponding infinite dimensional Lie groups, to the extent that they even exist, have a much more complicated representation theory.

So now I can try to answer your question. Operator algebras were more or less invented in order to provide a nice model for quantum mechanics. The nice property that this model has - namely, that there is only one realization of it up to unitary equivalence - is no longer true in QFT. So one (implicit) goal of a lot of work in QFT is to cope with this situation and search for better foundations. I have no idea if C*-algebras are the best or most modern way to think about QFT - probably not - but a good place to start for a student is to learn the Stone-von-Neumann theorem in some reasonable generality since we can blame a lot of the difficulty of QFT on its absence.