Consider a generator with zero resistance as in your word figure of sect. 2: there can be no net E field in the generator wire yet we know from Maxwell that a non-conservative E field is generated - again as you say.
In the wire, in order to reconcile this paradox, an equal and opposite E field is generated in the generator wire which is conservative, yielding net zero E field. Further, since the circulation of a conservative field has to be zero it follows immediately that the E field in the external resistor R is 100% conservative and equal but opposite to the conservative E field in the generator..
Call the non-conservative field ## \bf E_m ## and the conservative field ## \bf E_s ##. The total E field is just ## \bf E_m + \bf E_s ## vectorially, everywhere around the circuit.
The relations are then
## emf = \int \bf E_m \cdot \bf dl ## (integration thru all the wire);
since ## \oint \bf E_s \cdot \bf dl = 0 ## it also follows that
## \oint \bf E \cdot \bf dl = emf ## numerically. Again, as you state.
Whereas
voltage V = ## \int \bf E_s \cdot \bf dl ##. Circulation of this integral follows Kirchhoff's voltage law ## \Sigma \Delta V = 0 ##.
If there is resistance in the generator wire then the ## E_s ## field is reduced by the amount ## iR_{int} ## where i = current.
## \oint \bf E_s \cdot \bf dl = 0 ## still of course. Current i is consequently reduced by the internal wire resistance to ## emf/(R + R_{int}). ##. emf is unaffected.
I recommend my blog
https://www.physicsforums.com/threads/how-to-recognize-split-electric-fields-comments.984872/
for further study.