Confusion with resistor and diode voltages when heated in a series circuit
In circuit A, let's rename the 1k resistor R1. The voltage across R1 can be calculated using the resistor divider formula as follows:

V = 10 * R1 / (10k + R1)

So that is why it is a bit confusing. R1 is on the top and bottom. But it works out mathematically that as R1 goes up, V goes up. And as R1 goes down, V goes down. You can graph it or plug in a few values to see this. Or solve for R1.

Note that resistors can have positive or negative temperature coeficient. So it is not a given that heating up R1 will make the resistance go up. It may remain unchanged, or even go down.

Now let's talk about circuit B. For ordinary silicon diodes, with constant current, the forward voltage goes down as temperature goes up. But it is not correct to think of the diode as having DC resistance. Diodes don't behave like resistors. The way the diode behaves, when current is flowing through it, is that the voltage is relatively constant, only changing a little bit as the current changes. That is just how diodes behave.

So what happens is, as the temperature goes up, the voltage of the diode goes down, and the current will go up slightly. But this increase in current will not be anywhere near enough to make up for the voltage decrease due to temperature change. So the temperature effect wins, and the diode voltage goes down.