Rigorous argument that $4$-current transforms covariantly as $4$-vector

Question

I have some troubles to follow the chain of reasoning in comments below this explanation by Luboš Motl that entropy density (in nonrelativistic picture a scalar valued object) can be promoted in relativistic framework to covariantly transforming $4$-vector object. Motl wrote:

[...] the nonrelativistic entropy density, just like charge density, must be extended to a 4-vector "current" in any relativistic theory. It is because the total charge is actually a charge crossing a 3-surface within a 4-spacetime, so the integral has the measure dV^mu, and the mu index must be canceled by the same index from the vectory, so it must be a j^mu.

So it seems that the presented argumentation mimics the reasoning how to justify that the $4$-current $j^{\mu}=(c\rho, \vec{j}) $ transforms covariantly as a $4$-vector.

Question: But I cannot follow several parts in given argumentation chain; could maybe somebody elaborate it? It looks like a kind of "units fitting" argument but I'm not sure how to extract from this precisely the desired conclusion.

Thoughts (maybe this already resolves (parts of) my problem): Let take a spacelike $3$-surface $M$ (ie playing the role of "spatial part" of the spacetime) contained as a hypersurface slice in the $4$-spacetime.
To my knowledge the quantity "total charge" with resp $M$ is given by $Q_M = \int_M j^{\mu} \cdot \vec{n}_M dM$ as "surface integral" (...at least that how I understood the part that "total charge is actually a charge crossing a 3-surface within a 4-spacetime"). Is this correct? If not, ideas what was meant there?

Next is written that "the integral has measure $dV^{\mu}$, so that of the $4$-spacetime. I'm not sure if I understand that part. Isn't this just specialization of the more general statement that if we have a space $V$ of dimension say $d$ and $M \subset V$ is a hypersurface, and $\vec{F}:V \to TV$ is a vector field, then the surface integral with resp $M$ $\int_M \vec{F} \cdot \vec{n}_M dM$ should have unit/dimension of $V$? If not, what is else meant in the quoted setence?

Lastly, from previous considerations is concluded that "mu index must be canceled by the same index from the vectory, so it must be a j^mu." (...so transforms as $4$-vector. What is meant there by this "cancelation argument" precisely?

Rmks: I know different verification how to check 'by hand' from other considerations that the $4$-current $j^{\mu}$ transforms as $4$-vector covariantly, but to goal of this question is to understand the argumentation techniques (...which I not saw before, that's why I find them rather interesting) in quoted argumentation.
I have background in mathematics, so sorry if for trained physicists they appear everydays routine.

Answer 1

You're overthinking this, all that the argument is supposed to be is this:

If you know that charge is the integral of $j^\mu$ over a spacelike hypersurface, then for this to make sense $j^\mu$ must be a four-vector so that the expression $$ Q = \int_V j^\mu \mathrm{d}V_\mu \tag{1}$$ over a 3-surface/volume $V$ is well-defined. The "measure" $\mathrm{d}V_\mu$ is just one of the many ways to write the 3-surface (volume) element of the hypersurface. The two $\mu$s "cancel" each other by the implied summation so that the result is a scalar.

Whether this is a convincing argument depends on how one arrived at the belief that eq. (1) holds without somehow already arguing that $j^\mu$ is a 4-vector.