Consider a Euclidean CFT in radial quantisation, and let $S$ be the unit sphere centred on the origin. The state-operator correspondence says that any state $\Psi_S$ living on $S$ can be prepared by a path integral with an insertion only at the origin. It is proved as follows (see Sec 4.6 in Tong):
Let $S_r$ be the sphere of radius $r$ centred on the origin. Then we can evolve the state $\Psi_S$ radially to get some state $\Psi_r$ living on $S_r$. This evolution can be written as a path integral on the annulus $r\leq |\mathbf{x}|\leq 1$:
\begin{align}
\tag{1}\Psi_S[\phi_S] &= \int D\phi_{r} \space\Psi_r[\phi_r] \int_{\phi|_{S_r}=\phi_r}^{\phi|_S=\phi_S}D\phi e^{-S} \\
\tag{2}&=\int_{\phi|_S = \phi_S} D\phi \space\space\Psi_{S_r}\big[\phi|_{S_r}\big]\space e^{-S}.
\end{align}
(I'm working in the wavefunctional picture, where $\Psi_S$ is a functional of the field configuration $\phi_S$ on $S$, and likewise for $\Psi_r$. The second line follows since integrating over $\phi_r$ removes the inner boundary condition.)
Eq (2) tells us that any state $\Psi_S$ can be prepared by a path integral on the annulus $r\leq|\mathbf{x}|\leq 1$ with some appropriate insertion $\Psi_r[\phi|_{S_r}]$ on the inner boundary.
Now take $r\to 0$, so the inner boundary shrinks to the origin. Hence we conclude that $\Psi_S$ can be prepared by an insertion at the origin.
Question 1: How exactly is the $r\to 0$ limit defined? Each $\Psi_r$ is a functional depending on $\phi|_{S_r}$, whereas in the $r\to 0$ limit we generally expect to obtain an insertion depending on not just $\phi(0)$ but also its derivatives. I think the existence of an appropriate limit is the exact content of what people call the "state-operator correspondence", but I can't find any reference addressing this.
Question 2: Why does the QFT need to be conformal for the above to work? Even without conformal symmetry, the path integral still defines some map from states on $S$ to states on $S_r$. Then we can take $r\to 0$ as before and the rest of the argument seems to work fine, showing that all states can be prepared by an insertion at the origin. In other words, I'm claiming that radial quantisation is perfectly well-defined in an arbitrary QFT. Unlike in CFT, in general the 1-parameter family of evolution maps from $S$ to $S_r$ will not be the exponential of some conserved charge, but I don't see why this would pose a problem to the above argument.
I'm asking these two questions together because I suspect their answers might be related. For example, perhaps the limiting procedure in the state-operator correspondence is well-defined precisely when the theory is conformal.
EDIT: another confusing point is that usually in QFT operators have to be smeared: they're not well-defined "at a point". So I don't know whether the state-operator correspondence is even well-defined, at least the way it's normally written.
Best Answer
It is not true that any state on the sphere can be prepared by a local operator at the origin. For example, take the state $|\Psi\rangle=\phi(x)\phi(-x)|0\rangle$, defined on the unit sphere, where $0<|x|<1$ and $\phi$ is some local scalar operator. If it were true that $|\Psi\rangle=\mathcal{O}(0)|0\rangle$ for some local operator $\mathcal{O}$, then the correlation function $\langle 0|\mathcal{O}'(y)|\Psi\rangle$, where $\mathcal{O}'$ is a local operator, would be regular for all $|y|>0$ (the defintion $\langle 0|\mathcal{O}'(y)|\Psi\rangle$ makes sense only for $|y|>1$ but since correlation functions are analytic, this is enough to ask questions also at |y|<1). But of course we know that $\langle 0|\mathcal{O}'(y)|\Psi\rangle=\langle 0|\mathcal{O}'(y)\phi(x)\phi(-x)|0\rangle$ has singularities at $y=\pm x$ and no singularity at $y=0$.
As Connor Behan points out in the comments only the dilatation eigenstates can be represented by local operators at a point. For dilatation eigenstates, there is no limit to be taken as they evolve trivially. To make the argument precise, one needs to define what you mean by a local operator.
The objection raised by the OP to Connor Behan's comment is addressed by the following observation. The dilatation eigenstates do not span the Hilbert space, they only span a dense subset. In other words, to reproduce a completely general state, one needs to take infinite sums of dilatation eigenstates. These sums will converge in the Hilbert space, but not in the space of local operators (by the argument in the beginning of this answer).