Electric charge is a "special" kind of physical property because it corresponds to a very simple physical effect. But that's not true of most physical properties. The lepton number doesn't have any force associated with it, the way electric charge does, because it's not a coupling constant.
Lepton number is just a mathematical expression of what it means to be a lepton. The quantum fields which correspond to the particles we call leptons (electron, muon, tau, and their corresponding neutrinos) each have a lepton number of 1, and the fields corresponding to their antiparticles each have a lepton number of -1. Lepton number is considered to be a useful property because it is conserved in all observed reactions.
In general the theory with nontrivial chiral structure isn't invariant under chiral transformations of the fermion field; corresponding phenomena is called anomaly, and it leads to nonconservation of corresponding current. Formally it is related to the fact that there is no way to define bot gauge invarian and chiral invariant regularization for such class of diagrams. Particularly, if we have $U(1)$ global nonchiral symmetry for theory with fermions, but fermions interact through chiral theory, then this $U(1)$ symmetry becomes broken. For example, in SM the baryon number is defined through corresponding global nonchiral symmetry, and to break it we need to look for chiral gauge symmetry group. SM local symmetry group is based on $SU_{c}(3)\times SU_{L}(2)\times U_{Y}(1)$, from which the only chiral group is $SU_{L}(2)$. So that the only anomaly is $U(1)SU_{L}(2)^2$: we have that
$$
\tag 1 \partial_{\mu}J^{\mu}_{B} = \frac{3g_{EW}}{16\pi^2}F_{a}^{SU(2)}\tilde{F}_{a}^{SU(2)}
$$
Absolutely analogical thing is for lepton current:
$$
\tag 2 \sum_{l}\partial_{\mu}J^{\mu}_{l} = \frac{3g_{EW}}{16\pi^2}F_{a}^{SU(2)}\tilde{F}_{a}^{SU(2)}
$$
Why this nonconservation is nonperturbative? The reason is that $F\tilde{F}$ can be expressed as full derivative, $F\tilde{F} = \partial K$, and this means that the contribution of corresponding correlator in Feynman diagrams is exactly zero in all orders (look here for details). But in fact such correlator is nonzero due to nontrivial topology of the $SU(2)$ group. Corresponding comfigrations for which this term is not zero called (in electroweak theory) instanton-like configurations (pure instantons are forbidden). The other fact (it is more important here) which relates anomalous current nonconservation to nonperturbative effects is that the anomaly equations $(1),(2)$ are one-loop exact: we need to compute only triangle diagram for getting exact in all orders Eqs. $(1),(2)$. Corresponding theorem was proved by Adler and Bardeen.
Which Goldstone modes you discuss? There can't be spontaneous breaking of symmetries such as baryonic and lepronic symmetry in SM. This is (again) nonperturbative result which was proved by Witten. The breaking of baryon and lepton numbers symmetries is explicit, not spontaneous.
Best Answer
Let me address your questions one by one.
The SM Lagrangian is invariant under the fermion transformations, $$ \psi \to e^{iL\theta}\psi $$ where $L$ is assigned such that $e^-$, $\mu^-$ and $\tau^-$ leptons and lepton-neutrinos have $L=1$, whilst their antiparticles have $L=-1$, and everything else has $L=0$. This global $U(1)$ symmetry corresponds to lepton number conservation - $L$ is what we call lepton number. Lepton number is by construction the Abelian charge corresponding to that $U(1)$ global symmetry.
The lepton $U(1)$ global symmetry is accidental. We simply cannot write a gauge invariant, renormalizable operator in our Lagrangian that breaks conservation of lepton number. There is no reason to expect that physics beyond the SM respects lepton number conservation.
The electric charge $U(1)_{em}$ local symmetry was a principle on which the SM was built. The SM would not be renormalizable if $U(1)_{em}$ was explicitly or anomalously broken. It would be catastrophic if $U(1)_{em}$ were broken.
They are the conserved charges associated with different $U(1)$ symmetries. In general, when you have multiple $U(1)$ symmetries, charge assignment is somewhat arbitrary, since one can pick different linear combinations of the original $U(1)$ generators as the symmetries. In this case, however, the $U(1)_{em}$ is local, so there is no mixing with the global lepton number $U(1)$.