Why do charges not lose potential as they travel through the circuit before reaching a resistor?
The resistance in a real circuit would be far higher than the resistance of the lead wires, so the bulk of the electric field would be seen across the resistor.

Now we need to interject some reality.  In a schematic, the resistor is a lumped element defined by properties at its terminals.  IRL, however, it is a volume of resistive material.  Two of the three dimensions of this volume are typically width, and the third is length.  

If you take some fraction of that length, it will have a smaller electric field across it than the whole device does.  If one side is ground, different parts will show different potentials depending on how close they are to the non-ground terminal.  This is, in fact, how analog potentiometers work--the center terminal moves from one end of the fixed resistor to another, changing its potential depending on its position.

So the electric field is distributed around the circuit.  The trick is that most of it is distributed across the resistive element, with very little distributed around the lead wires.