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If the Hardt-Simon foliation was just a lamination and not a foliation, then you might have to be more careful, since there are laminations which don't admit transverse measures, and so it might not be clear that any function parametrizes its leaves. So it's good to be careful about these issues, but luckily here it causes no problems. (ps - thx!)
Usually when people say "level set of a function of least gradient" they mean "boundary of a superlevel set of a function with least gradient", but the somewhat stronger thing that you're asking for can also be arranged. Let me quickly update my answer @LeoMoos.
I think by looking at pictures of half-catenoids it is pretty easy to geometrically see why this is a counterexample. As $\varepsilon \to 0$, the neck gets thinner and thinner, while the wider part gets fatter and fatter. The Harnack constant measures the ratio of the fatness to the thinness, and as $\varepsilon \to 0$, the neck shrinks to a point while the wide part becomes infinitely wide.
@user88544 The point I was trying to make is that on the linear level, Harnack's inequality depends on the ellipticity, so on the quasilinear level, the same thing had better be true. However, this was kind of hazy and worded confusingly by me, so I am sorry for that. I have added a more explicit counterexample. The Harnack inequalities for the catenoids blow up as $\varepsilon \to 0$.
Could they possibly mean that the map is to $L^4(M, \mathbb R^n)$ where $N \subset \mathbb R^n$? This is a linear space, and contains the tangent spaces of $N$ as subspaces.
