Electric stove heating element and switches
From one perspective I have learned the more resistance in say a conductor the hotter the conductor itself gets. The smaller wires get hotter as thicker wire with respect to the same voltage applied to both.


Don't confuse power and temperature. Temperature will stabilise when heat last to surroundings = power in.
Power, \$ P = \frac {V^2} R\$ so as resistance decreases power increases. Thicker wires will dissipate more power.


I am certain that the heating element hardly has any resistance in that sense.

Rearranging the formula we can calculate the resistance. For example, a 1000 W heater element on a 230 V supply, \$ R =\frac {V^2} P = \frac {230^2} {1000} = 53\ \Omega \$. Whether you think that is &quot;hardly any&quot; I can't say.

Or if the lower resistance heating element won't get as hot as the higher resistance element.

Try putting different values into the equation.

With respect to the control switches, if they control voltage by means of having their own some sort of metallic variable resistors which in turn must get hot and get old. And if this varying voltage determines the amount of current to the heating element and not the heating element itself.


Figure 1. A typical bimetallic strip cooker power controller. Read about its operation in another answer. Image source unknown.
Most heaters use on-off control. If you look at a ceramic hob you can see the heater turn on and off with the duty cycle varying from about 5% to 100% as the control knob is adjusted from min to max. Note that these control power, not temperature. The temperature will depend on what is put on the hob. No pan -&gt; very hot. Pan full of water, not so hot.

I am also wondering ... what happens to resistance as the heater gets old.

Hot wire is more fragile than cold and can get damaged. Any narrow spots in the wire have to carry the same current as the rest of the wire and due to their higher resistance per unit length in this area will run hotter and tend to fail at that point.