Now, if we apply this logic to a circuit, it seems to me that even if
there is 0 resistance, charges starting from the positive terminal and
going to the negative will have lost 12V
By Ohm's law (I suppose you accept Ohm's law since you mention resistance), the voltage across a zero ohm resistance is zero for any finite current through.
If you accept this, it follows, (by applying logic), that it isn't the case that charge flowing through the external circuit "will have lost 12V", since there is no voltage across a zero ohm resistance to 'lose'.
Now, if the 12V battery is a physical battery, i.e., can only supply finite current into a zero ohm load, there is no contradiction; the internal resistance of the battery will 'drop' the 12V (the battery will get hot) and the terminal voltage will be zero (the short circuit will not get hot).
If, by 12V battery, you mean an ideal 12V voltage source, then you have a genuine contradiction, i.e., 12V = 0V. This is why, as is well known in ideal circuit theory, ideal voltage sources cannot be connected in parallel with ideal wires.
It is not an issue of the field being conservative or not. Ultimately, Kirchhoff's laws are about the relationship between branch currents and node voltages in a network of lumped circuit elements. If you define three kinds of branch elements denoted by $R,C,L$ using the relationships $v=Ri$, $i=C\frac{dv}{dt}$, and $v=L\frac{di}{dt}$, respectively, then you may freely use Kirchhoff's current and voltage laws. These defining relationships between voltage and current are idealization and simplification not just for an inductor but also for a capacitor and resistor, as well. In the case of the inductor we ignore all fields outside the coil, and if we cannot because we have an inductive transformer then we include that part explicitly by defining a two-port with a pair of equations, such as $v_1=L_{11}\frac{di_1}{dt}+L_{12}\frac{di_2}{dt}$ and $v_2=L_{12}\frac{di_1}{dt}+L_{22}\frac{di_2}{dt}$, and a similar set of equations if you need more ports than two. If the capacitor is physically large then we may encounter problems with the current continuity law and will not be able to neglect the displacement current.
Note too that in no sense one could claim that the fields of a voltage or current generator are "conservative", not even for a battery: electrochemistry is not electrostatics. Somewhere, somehow you must impose a phenomenon that is outside of electricity or magnetism. Instead we postulate that certain node pairs have a predefined voltage history, and a given branch has a predefined current history independently of the rest of the circuit and thus represent a voltage or a current source, resp. In other words sources are time dependent boundary conditions. This way as you go around in a loop you must always get 0 voltage, no conservative field is needed. At the next level of abstraction you only need that in an arbitrary loop at any instant every connecting wire the current must be the same. And assuming linear superposition you can derive that the sum of branch currents at any node must be zero. So then the only questions is whether a loop is physically small enough so that the current uniformity holds. Once you have picked the defining lumped element equations between $v$ and $i$ you may say that KVL and KIL have more to do with network topology than actual physics.
Best Answer
Imagine a free-standing battery (not connected to any wires) and take a closed loop through the battery, out one terminal, and back in the other terminal. The total work done in moving a test charge around that loop must vanish. For this to happen, the change in electric potential outside of the battery must equal the negative of the EMF change within the battery. $$\int_\mathrm{outside}\vec{E}\cdot{\mathrm{d} \vec{\ell}} + \mathcal{E} = 0$$
Update after comments
Work is well defined as the integral of force over distance. The relationship between work and energy is more subtle. One needs to carefully define what the system in question is. We also have to recognize that potential energy is the energy associated with the configuration of a system of interacting entities. One is on thin ice if one refers to "the potential energy of a particle". Particles do not have potential energy. The system comprising the particle and something with which it interacts has potential energy. (A ball does not have potential energy. The earth-ball system has potential energy.) I'll review the background on this, with apologies if the background is already well-understood.
Once a system is defined, energy can be added to the system by an external force which can do external work on the system. Work is one way energy can be added. Heat is another, but we'll mostly ignore heat and thermal energy. Generally, $$W_\mathrm{external} = \Delta E$$ where $E$ is the total energy of the system. External work causes energy to be added to the system, but once inside that energy could be potential, kinetic, thermal, chemical ...
Potential energy is defined to be the negative of the work done by forces internal to the system:$$W_\mathrm{internal} = -\Delta PE$$
Now our system. Let's take it to be the wire, the battery terminal, the conductors inside the battery, but not the chemicals and processes that generate the "chemical force". The chemical processes are a source of energy, so we'll take it to be outside of our system. The work done by the chemical processes are external, and do external work on the charge carriers $$W_\mathrm{external}=q\mathcal{E}$$ But the internal voltage due to the separated charges within the battery, also do work, but this work is internal to our system, and thus changes the potential energy of the system $$W_\mathrm{interal} = -\Delta PE = -qV$$ but recall $$W_\mathrm{external} = \Delta E = \Delta PE = -W_\mathrm{internal}$$ (ignoring stores of energy other than potential energy within the system). Finally $$q\mathcal{E}=qV$$ $$\mathcal{E} = V$$