How does mass-energy equivalence work differently in nuclear decay where there is nucleus excitation? I feel that I have a misconception somewhere in my understanding of quantum physics and nuclear decay.
In fission, when the products have less mass in total than the reactant, the excess is understood to have converted completely into kinetic energy possessed by the reactants who travel off in opposite directions at extremely high speeds.
Similarly, the mass defect between the mass of a Tritium nucleus and its constituents, 1 proton and 2 neutrons, is representative of the binding energy per nucleon of Tritium.
In decay, there is gamma decay, or the emission of electromagnetic radiation in the form of high frequencied photons. These occur when a nucleus is in an excited state following either alpha or beta decay.
In the standard beta plus decay, a proton decays into a neutron, releasing a positron and electron neutrino in the process. The mass of a proton is lighter than that of a neutron, and it's part of the reason why beta plus decay is non-spontaneous and requires external agents for it occur.
However, in beta minus decay, a neutron decays into a proton, releasing an electron and electron antineutrino in the process. Now, there is mass excess in this decay. Is this mass excess converted into kinetic energy to be possessed by the electron and also the electron antineutrino? Also, in a bigger picture, Cobalt-60 decays via the same process into Nickel-60 releasing 2 gamma rays in the process due to nucleus excitement. How can we tell if there will be nucleus excitement due to alpha or beta decay, and where does the extra energy from these gamma rays come from? 
Also, I read about the formation of a Carbon-12 atom from 3 alpha particles. But alpha particles don't have electrons, so how is a Carbon-12 atom formed from 3 alpha-particles? It's similar to the decay of Carbon-14 to Nitrogen-14, where are the extra electrons coming from? I feel that I don't entirely understand the quantum energy level concept of electrons.
 A: There are more questions there.

part of the reason why beta plus decay is non-spontaneous and requires external agents for it occur.

Forget the concept of external agents in beta decay, unless you address a very specific topics. It is an action from nucleus to nucleus, you cannot separate single nucleons.


*

*The beta decay of the nucleus is not described as change of one single proton or neutron, but rather the complete transition from one complex state to another - $\left<\psi_{A_{Z\pm1}}\psi_\nu\phi_e|H_{int}|\psi_{A_Z}\right>$.
Via some interaction $H_{int}$. Read from right to left.
Now realize, that the $ m(A_{Z\pm 1}) \lt m(A_Z) $ and so there is some energy available (on behalf of the missing mass) to turn into kinetic energy, as you say and expect.

*Until this moment, we assumed, that the beta transition was always from a ground state to a ground state. But - nuclei usually have some excited states - one can imagine them as a vibration, rotation or something more quantum mechanical. And we know some ways how to push (or create) a nucleus into such a state. Very usually, these states just decay by gamma decay (as you said).
Such an excited state has higher mass. $m(A_Z^*) \gt m(A_Z^{gs})$. If the excited state has energy e.g. $E^*=900\rm\,keV$, it is $900\rm\,keV$ more heavy and it can/will irradiate $900\rm\,keV$ photon after a while.

*Now the connection between $\beta$ and $\gamma$: some beta decays lead not (only) to the ground state of the daughter, but also (or even uniquely) to the excited states. WHY? $\psi_{A_{Z\pm1}^{g.s.}}$ can be very different from  $\psi_{A_Z}$ and there could exist several more similar excited states like $\psi_{A_{Z\pm1}^{*}}$. From the energy budget: a bit less $(E_{budget}-E^*$) will go to the electron-(anti)neutrino pair, and the rest $E^*$ will be brought away by a gamma deexcitation in the next moment.

But alpha particles don't have electrons, so how is a Carbon-12 atom formed



*Don't care about electrons. Their binding energy is like $13\rm\,keV$, while you play with $8000\rm\,keV$ per nucleon. Sometimes you can miss an electron, but this is no problem, electrons are everywhere, brush your hair with plastic and you create an electron Armageddon. 


As a bonus I add an isobar levels diagram, $\beta$ decays can proceed following the diagonal arrows. The $y$ axis is an excitation energy but also the mass of the nuclear state.

