Why is the electric field highest in regions of highest resistance? I know Kirchhoff's law and that current must be constant. But what exactly is the phenomenon, from first principles, that causes the electric field to be stronger in a region with lower carrier mobility/concentration. 
I thought it might be repulsion from electrons  that pushes the electrons in front ahead (as charge instantaneously accumulates) but I can't really wrap my head around how exactly more work is done to move a charge through a higher resistance.
 A: Are you speaking specifically about currents in wires?  If you look at (the simplest version of) Ohm's law you will see that $\mathbf{J} = \sigma \mathbf{E}$, where $\sigma$ is the conductivity (technically it's a tensor, but we'll assume a constant scalar for now).  In this case, $\eta$ = $\sigma^{-1}$, which is the resistivity.  Thus, we can show that $\mathbf{E} = \eta \mathbf{J}$.
So if the current, $\mathbf{J}$, is constant, then when the resistance (= resistivity per unit length) is high, the electric field $\mathbf{E}$ should be high, if this simple approximation of Ohm's law holds.
Side Note
Generally, it is true that the electric field and current density are coupled to each other, but Ohm's law can be much more complicated.  So you should not blindly assume that when the resistance is high, the electric field will be high because of this relationship.  I make this point because in a plasma, the resistance is often supplied by the electric fields from electromagnetic and electrostatic waves acting as an effective "drag force" on the particles that are attempting to stream relative to each other.  
Thus, in this case a larger resistance to flow is caused by the electric fields having a larger impact on the particles, not a larger electric field caused by a larger resistivity.
A: $E = - \frac{\mathrm dV}{\mathrm dx}$
means that when voltage changes rapidly with distance inside the resistor, the electric field will be high.  You could have two, cylindrical, 1000 ohm resistors, one of which is 5 mm long and the other is 50 mm long.  If they are in the same circuit (carrying the same current), the 5 mm resistor will contain the higher electric field because the voltage changes more rapidly across it.
A: I believe what occurs is that surface charges accumulate on the surface of the resistor, preferentially toward the side with the negative potential (assuming the charge carrier is negative), inducing an electric field in that direction. And the gradient of surface charge density would be higher in regions of higher resistance.
