Voltage: $V_{AB}=\int_{AB} \mathbf E \cdot \mathrm{d}\mathbf l$, where the electric field is produced by charge buildup. This is different from potential difference, as potential difference is path-independent. Also, the sign is flipped (as opposed to potential), as most circuit components cause voltage drops and it is convenient to make their voltage positive.
EMF: work done by external force to move a unit charge across some length of wire
$$ \varepsilon = \frac{1}{q}\oint \mathbf{F} \cdot \mathrm{d} \mathbf{l} $$
Current is defined as $I=\mathrm{d}Q/\mathrm{d}t=\iint \mathbf J \cdot \mathrm{d}\mathbf A$
Each electron has a uniform probability of colliding and randomising the velocity, proportional to time (assuming constant drift velocity, which happens to be valid due to QM effects). $\mathbf F$ is the lorentz force.
$$ \frac{d\left<\mathbf p\right>}{dt}=-\frac{\left<\mathbf p\right>}{\tau} + \mathbf F=-\frac{\left<\mathbf p\right>}{\tau} + q\mathbf E $$
At steady state,
$$ \left < \mathbf p \right>=qE\tau $$
$$ \mathbf J=nq\left<\mathbf v\right>=\frac{nq^2\tau}{m}\mathbf E=\sigma \mathbf E $$
Usually we assume volume mobile charge density is close to infinite so the drift velocity is insignificant.
When calculating resistance through unconventional resistors, it is usually easier to first get $\mathbf{J}$, as $\mathbf{I}$ has to be conserved throughout the length of the resistor (in the absence of current sources/sinks) ($\nabla \cdot \mathbf J=-\frac{d\rho}{dt}=0$). ($\mathbf E$ can also be calculated quite easily, unless the conductivity changes, which will result in a net charge density within the volume). Then, $V$ can be attained through the integration of $\mathbf{E}$ (from Ohm’s law) throughout the length. Finally divide $V$ by $I$ to get the resistance.
At steady state, divergence of J is 0 as there is no change in charge density anywhere. Hence, the divergence of the electric field must be 0. This means that all net charge must lie on the surface of the wire, or any other discontinuous areas. The buildup of surface charge density is what creates the electric field (together with the batteries) that creates a current.
We can ignore these charges in the lumped element model as thin wires have negligible capacitance. The voltage difference must be negligible as we assume they have negligible resistance as well.