Let some field theory be described by the Lagrangian density ${\cal L}$ on spacetime. Noether's first theorem asserts that given a quasisymmetry $\hat{\delta}\phi$ there is a class of currents $j^\mu$ such that $$\partial_\mu j^\mu =E\hat{\delta}\phi\tag{1}$$ where $E$ are the equations of motion.
Two currents in the same class differ by a trivial current which can be either (1) a current that identically vanishes on-shell, (2) a current which is conserved even off-shell and (3) any combination of these.
Noether's second theorem states that when the quasisymmetry is local, i.e., parameterized by a function $f$, one such current associated to it, verifying (1), is some $S^\mu$ which vanishes on-shell $S^\mu\approx 0$. Therefore any other current in the class $j^\mu$ verifies
$$\partial_\mu(j^\mu - S^\mu)=0\Longrightarrow j^\mu=S^\mu+\partial_\nu k^{[\mu\nu]}\tag{2}.$$
In this paper by G. Barnich & F. Brandt the authors say that this gives rise to a "Noether charge puzzle":
Note that the superpotential is completely arbitrary because it drops out of (1.1) [Eq. (1) of this post] owing to $\partial_\mu\partial_\nu k^{[\nu\mu]}=0$. This implies that the Noether charge corresponding $\delta_f$ is undefined because it is given by the surface integral of an arbitrary $(n-2)$ form.
How the same problem does not happen for a global symmetry for which Noether's second theorem does not apply? I mean, the current class of such symmetry is not the trivial one anymore. Still, if $j^\mu$ is a current in the class we can always add some $\partial_\nu k^{[\mu\nu]}$. How is this any different than the local case?
More importantly if we define the Noether charge by integrating $j^\mu$ over a Cauchy surface $\Sigma$ is the charge, in the global case, well-defined? Because I see the same issue taking place in the global case. Let $j^\mu$ be a current in the class. We get another one by adding $\partial_\nu k^{[\mu\nu]}$, then the charge changes by a boundary term at $\partial \Sigma$.
