$$V_{ab}=-\int_b^a\vec{E}*\vec{dl}=\int_a^b\vec{E}*\vec{dl}$$
By Kirchhoff's loop rule, the sum of the potential drops across a closed loop is zero. Going in the direction of conventional current across a battery (negative to minus, thus in the opposite direction of $\vec{E}$) we have a positive voltage term. Subsequently, resistors must have potential drops to make the loop sum to zero. Therefore, for resistors our conventional current will cross resistors going in the direction of the electric field. Instead of gaining potential going across a resistor, our conventional current is having work done on it by the electric field across the resistor and thus having a potential drop in the loop.
My main question
So I've read a lot on stackexchange and quora about beliefs that resistors slow down electrons are overly-simplified, and even a post detailing the fact that resistors are materials with low free-charge carrier density. But I still haven't figured out how a resistor can create an electric field. I do understand it basically in the case of a battery being the result of chemical processes generating an EMF, but not in the case for resistors.
Previous beliefs I would say that the resistor causes a clogging of electrons on one side thus making the other side +ve. However, not only is this overly simplified but intuitively it seems that the side of electron clogging will be the side before entering the resistor thus by even that logic the resistor would form an electric field going opposite the flow of conventional current. In return, this resistor would not cause a potential drop but instead a raise.