Why does it imply that we can't localise
It doesn't. It says that linear methods cannot resolve past its limit.
Perfectly jointly localizing
Single-component signals can be localized perfectly in both time and frequency. For a pure sine, as long as sampling above Nyquist, two points are sufficient to perfectly recover its frequency, phase, and amplitude.
(Both above are of course limited by numeric precision). For variable frequency/amplitude, time-frequency reassignment works. Synchrosqueezing examples:
Rather than perfect, these examples are challenging and feature more than one "component", just to show that performance is good even with >1 component. SSQ_STFT should${}^1$ handle a (single) linear chirp perfectly, and SSQ_CWT exponential chirp, though there's fundamental limitations with finite inputs near boundaries (information discontinuity). Note, SSQ's performance on single-component is very impressive, even handling pure noise as the frequency (code in source post):
The mapping may be imperfect, it's also stretched (log-scaled) since it's CWT.
Another way to perfectly map a single-component AM-FM may be using unwindowed STFT with size-2 DFT, via the sine inverse solution. In addition to the Nyquist rate, the sampling rate is constrained by $f'(t)$, i.e. frequency derivative, e.g. we obviously can't losslessly map a jump from $f=2$ to $f=4$ that happens between samples, even if $f=4$ is below Nyquist. This constraint is formalized for synchrosqueezing (see linked post).
What's "single-component"? In short, it's controlled by the time-frequency kernel (e.g. wavelet/kernel), or the resolution of the linear method, but basically a low-frequency amplitude modulating a relatively high variable-frequency carrier is single-component - some details.
1: I'm not certain, I've not studied the math, but it seems very likely in continuous time, else near-perfect. What's for certain, the Heisenberg limit goes out the window for single-component.
"Resolution" means "ability to resolve"
If there's one sine, there's nothing to resolve. The Fourier transform is an imperfect tool for handling one sine. It is, however, a perfect (in a sense) linear tool for >1 sines.
The FT works by taking an inner product (similarity/dot product) of every possible sine against the input. What one learns when there's two sines, is that traditional limitations kick in if the sines aren't infinitely long - i.e. observation interval. With DFT, one can observe this by taking shorter and shorter durations of same frequency sines - the "correlation machine" that is the DFT will eventually fail to produce two distinct peaks.
Localizing general signals
Firstly, localization is non-unique:
$$
\cos(A)\cos(B) = .5[\cos(A+B) + \cos (A - B)]
$$
Second, not every signal has a meaningful AM-FM representation - e.g. unit impulse, rectangle function.
This aside, AM-FM localization of general signals amounts to solving the cocktail party problem - i.e. unmixing. Linear methods are incapable of perfectly (or, most of the time, even satisfactorily) unmixing. Yet, fundamentally, it's critical to acknowledge, that the case here is not same as the physics uncertainty principle: here, we have perfect information (the signal is exactly determined) - in physics, the problem is that we can't measure exactly to begin with. Obviously every real-world signal is subject to this limitation (plus general noise which far outweighs the quantum limit), but that's separate from the mathematical problem.
Thus, nothing says perfect unmixing is impossible. Except, noise (mathematically), due to non-uniqueness.
The most successful methods for joint localizers are deep neural networks: they have excellent performance in e.g. voice separation, meaning their latent representations necessarily break the Heisenberg limit. Obtaining AM-FM mappings from such outputs, or from latent representations, however, is another story - the separated voices are still very much multi-component.
Proof idea: "Linear = Heisenberg-bound"
For any sequence of operations, as long as the net-operation is linear
$$
\text{op}(c_0x_0 + c_1x_1) = c_0\text{op}(x_0) + c_1\text{op}(x_1)
$$
the operation has a Fourier transform${^2}$. Then, it suffices to prove that there cannot be a function $x(t)$ that is time-limited and also band-limited, i.e. $=0$ outside of $[a, b]$ in time and $[c, d]$ in freq. This is done in Wavelet Tour - full proof in screenshot. Hence, no linear operation can yield $a=b < \infty$ and $c=d<\infty$ at once.
2: I don't really know this. It makes sense to me, but I'm just regurgitating authoritative sources on the conclusion "Linear = Heisenberg-bound", which also may be a simplification; @Jazzmaniac:
The uncertainty principle is a property of bilinear (i.e. "inner product") time-frequency distributions.
The idea of using the invoked theorem as "proof" is also mine, I've no clue what the formal justification is. That all linear operators can be expressed as functions at all seems to be untrue (I'm unsure how to exactly interpret the Q&A), which'd make the statement definitively false.
Re: OP, what's "instantaneous frequency"?
Here I address the asker instead of just the title.
Can someone clarify this a bit?
referring to making sense of "instantaneous frequency". The field that studies this is called "time-frequency analysis":
OP's integral has the right idea concerning "instantaneous", but wrongly implemented. A universal definition of instantaneous frequency is
$$
f(t) = \frac{d\phi(t)}{dt}
$$
where $x(t) = A(t)e^{j\phi(t)}$, and $x(t)$ can be made real by adding complex conjugate (doesn't change $f(t)$). The $A, \phi$ immediately run into uniqueness problems, and such $f(t)$ has limited "meaningfulness" beyond single-component signals. But, it's great for making sense of inst freq: once understanding the rotational nature of complex numbers, it's simply the speed of a phasor's rotation. One can also define a non-vanishing $\text{lim}$, but differently.
