First, a misconception (or rather, an oversimplification). QFTs are local, but that does not mean they are insensitive to global properties. For example a free fermion with anti-periodic boundary conditions has a propagator $\psi(z)\psi(w)=1/(z-w)$ but if you choose periodic boundary conditions, the propagator becomes $\psi(z)\psi(w)=(1/2)(\sqrt{z/w}+\sqrt{w/z})/(z-w)$. The two propagators are different, and more generally most properties of the theory are different (e.g., the spectrum of $H$). The fact that the theory is local means that, as $z\to w$, the propagator is independent of the boundary conditions. But it does not mean that the full theory is. It clearly is not. In this sense, the statement
Quantum field theory (QFT) is supposed to be local, which in this context means that it should be defineable on any spacetime manifold with the given number of dimensions (four in this case) because they all look the same locally.
is very much not true. If you can define the same theory on two different manifolds, then local questions yield the same answer. But non-local ones (which most questions are, e.g. correlation functions, particle spectrum, etc.) do not. So it is perfectly conceivable that one of the theories is well-defined while the other is not -- these are two different theories after all.
To give a very specific example, consider the Standard Model. It is chiral, so it depends on the orientation of spacetime. The theory is well defined on orientable manifolds, but it is meaningless on non-orientable manifolds. You need an orientation in order to define the theory, because left-handed fields interact differently from right-handed ones; this statement is meaningless if you cannot distinguish left from right, e.g. if you are on a 4d Möbius strip.
Second, back to anomalies.
An anomaly is a property of the theory, i.e., of its degrees of freedom. It is not a property of a realization of the theory. What I mean by this is, the anomaly does not depend on background fields. And gravity (i.e., the manifold you put the theory on) is a specific choice of background field. So the anomaly does not depend on which manifold you put your theory on. It is intrinsic to the theory.
Consider e.g. the axial anomaly. This anomaly is the statement that, if you have $n$ chiral fermions and turn on a background $U(1)$ connection, then the partition function shifts as
$$
\delta \log Z=e^{i Q\nu},\qquad Q:=\sum_{i=1}^nq_i^3
$$
under a gauge transformation of this background connection. Here $q_i$ are the $U(1)$ charges of the fermions, and $\nu\sim\int F\wedge F$ is the instanton number.
The anomaly is $Q$, not the phase itself. If you choose a background field with zero instanton number, $\nu=0$, then there is no phase, yet the theory is still anomalous. With this choice of background, you cannot detect the anomaly, but the system is still anomalous.
In the most general case, you turn on background fields for all the symmetries (including Poincaré), or at least as many as possible. This means, arbitrary instanton number for $U(1)$ symmetries, arbitrary manifolds for gravity (including non-orientable if possible), arbitrary magnetic fluxes for center symmetries, etc.
The more general your background, the more anomalies you can detect. If you choose simpler backgrounds, you can detect fewer anomalies. Which is fine, but it does not mean the anomaly is not there; it just means you didn't work hard enough, and you missed some. This is precisely what people are doing nowadays: they are taking old systems and turning on more general backgrounds, thus detecting more general anomalies, and therefore obtaining more information about the QFT they were interested in.
For example, if you reconsider the $U(1)$ symmetry from before, but now also turn on a gravitational background, you observe that the anomalous shift is
$$
\delta \log Z=e^{i Q\nu+iQ'\sigma},\qquad Q':=\sum_{i=1}^nq_i
$$
where $\sigma\sim\int R\wedge R$. So we see that $U(1)$ may have two different anomalies (and, if you wished to gauge it, you'd need to make sure they both vanish). You can only see both anomalies if you choose sufficiently non-trivial backgrounds which, in this case, means both non-zero instanton number and non-zero signature. If you had a simpler background, you'd be sensitive to less anomaly information.
Also, going back to the free fermion example, a circle with anti-periodic boundary conditions is null-bordant, with bordism e.g. a disk. The circle with periodic boundary conditions is not null-bordant. So a fermion with the latter boundary conditions has ill-defined partition function, according to the references. But as Witten points out, this is a rather mild issue. It just means that the partition function is defined up to a sign, and there is no canonical way to choose it. But this is fine: just choose whichever sign you want. This is "free parameter" of your anomalous theory. As a fun fact, this free choice is what gives rise to the two types of Type II String Theories. Namely, for one sign you get type A and for the other type B. (See arXiv:1911.11780 if you are interested in the details.)
Hope this helps.
