Applying Gauss law to a disk-like distribution In an paper titled: "Mass Distribution in Rotating Thin-Disk Galaxies According to Newtonian Dynamics" by Feng and Gallo:
https://www.mdpi.com/2075-4434/2/2/199/htm
the fourth paragraph of the Introduction reads:

Unlike the spherically symmetric mass distribution that generates the local gravitational force at a given radial position only depending upon the amount of mass within that radius, the gravitational force due to a thin-disk mass distribution can be influenced by matter, both inside and outside that radius. Thus, the mass distribution in a thin-disk galaxy cannot be determined simply by applying Keplerian dynamics, which relates the mass within a radial position to the local rotation speed.

While I can see how applying Gauss' law may not always be useful, I do not understand how the law itself isn't applicable for a disk-like distribution. How is there an influence of matter from both inside and outside the radius of interest?
 A: I think this is referring to Newton's shell theorem, which
can be derived from Gauss's law for gravity.
Newton's shell theorem says that the gravitational influence of any spherically symmetric mass distribution on a body external to that mass distribution, can be dealt with by assuming that all the mass is concentrated at the centre. It further notes that the gravitational field inside any spherically symmetric mass distribution is zero. That means if you want to know the gravitational field within a spherically symmetric mass distribution, you only need to consider the mass interior to that position and treat it as a point-like mass at the center.
Gauss's Law for gravity is
$$\oint \vec{g} \cdot d\vec{A} = -4\pi G M\, , $$
where $M$ is the mass enclosed by the surface.
If the mass is spherically symmetric then we can do the closed surface integral over a sphere and the spherical symmetry tells us that $\vec{g}$ must be of constant magnitude over that surface and that it is directed radially (inwards). This means that the scalar product in the integral is trivial and the LHS is just $4\pi r^2 g_r$ and we recover the Newton's shell theorem result.
Now consider a thin disk. What surface are you going to construct? There is an obvious symmetry in the disk plane which means that the gravitational field there will be of constant magnitude at a given radius and will be directed inwards. But what about above the disk plane? The field will not be of the same magnitude at a similar radius and it will also not be directed radially inwards. In fact, in the limit of an extremely large, uniform disk, if you move a little way out of the disk plane then the gravitational field will be pointed approximately directly back towards the disk plane. Thus the LHS of Gauss's law becomes non-trivial to evaluate.
Having said that: of course Gauss's Law is still true, it is just difficult to apply and a more tractable approach is to calculate the potential and take its gradient. The results for a uniform disk are surprisingly messy (see Krough, Ng & Snyder 1982).
But what about the claim that the mass exterior to some radius now has an influence. Well yes, for a non-spherically symmetric mass distribution that will be the case. If you remove all the mass from interior to some radius, all that Gauss's Law tells you is that since $M=0$, the integral on the LHS is also zero. But that does not tell you what the gravitational field is zero. It can be (and is) non-zero due to the non-spherically symmetric distribution of mass exterior to it. It is only for the spherically symmetric cases that the contributions from all parts exterior to the radius cancel out.
Further discussion of Feng & Gallo's ideas can be found at Astronomy SE.
