Why are potential drops concentrated at resistive elements? I do not understand why potential differences in the electric field in a circuit are concentrated at resistors when the potential, to my knowledge, is purely dependent upon location in the field and the magnitude of the charges. Why is the decrease in potential concentrated at resistive elements, rather than decreasing linearly with distance travelled through the circuit?
Similar questions have been asked already on this platform but none have alleviated my confusion.
 A: Lumped elements (including resistors) are idealized models which allow you to deal with electronics without solving Maxwell's equations. Of course, they don't exist in reality: the E field doesn't change abruptly next to a resistor, and there's no such thing as a pure resistor to begin with: it will have some inductance and stray capacitance as well.
Think about rigid bodies in mechanics for comparison: how could they generate any forces on collision if they don't deform? Well, we just assume they do, and it turns out that such simplified models are still useful in some cases.
Resistors work in the same way: if you can neglect all the "parasite" parameters in the electric circuit, you can pretend the field is really concentrated inside, and your model will still predict reality quite well.
A: That the potential drop across an ideal (resistivity $\rho = 0$) wire is zero is a consequence of Ohm's Law:
$$\mathbf{E} = \rho\,\mathbf{J} = \mathbf{0} \Rightarrow \Delta V = -\int{\mathbf{E}\cdot\,d\mathbf{l}} = 0.$$
($\mathbf{J}$ is current density, which, if you aren't familiar with it, is essentially a current per unit area.)
In a non-ideal circuit element, on the other hand, $\mathbf{E} \neq \mathbf{0}$ (for a nonzero current), making the potential drop nonzero. The most important case is that of a homogeneous cylindrical chunk of material (length $L$, cross-sectional area $A$). If we set up a uniform current density along the cylinder, the potential difference across it is
$$\Delta V = -\int{\rho\,\mathbf{J}\cdot{\,d\mathbf{l}}} = -\rho JL = -\frac{\rho L}{A} I.$$
That is, the voltage is proportional to the current. The constant of proportionality, as you probably know, is called the resistance.
It might also be helpful to you to think about what happens at the interface between two conductors of different resistivity. Consider two connected wires, one of resistivity $\rho_1$ and the other $\rho_2$, with $\rho_1 \neq \rho_2$. When the system is in steady state, the current density just to the left of the interface must match that just to the right-- otherwise the surface charge at the interface would change with time, as either more or less charge would be flowing in than flowing out, in a unit time. That is, $\mathbf{J}_1 = \mathbf{J}_2 \equiv \mathbf{J}$. Hence
$$\mathbf{E_2} - \mathbf{E_1} = (\rho_2 - \rho_1)\mathbf{J}.$$
Gauss's Law with a pillbox at the interface as the surface then yields
$$\sigma = \epsilon_0(E_2 - E_1) = \epsilon_0 J(\rho_2 - \rho_1),$$
where $\sigma$ is the surface charge density at the interface. So a surface charge accumulates at the boundary between materials of different resistivity in a circuit.
A: 
the potential, to my knowledge, is purely dependent upon location in the field and the magnitude of the charges

That is correct.
What happens is that when current is passed from a highly conductive material to a resistive material there are “surface” charges formed at the boundary between the two different materials. This charge distribution produces a greater E field in the resistive material, where the potential drop is sharper. The relationship between the resulting E fields and the charge distribution is as expected from Gauss’ law.
