How do we get velocity and position from motion with changing speed, and how do we apply kinematic equations to free fall?

Your arithmetic is right: 20 m in 2 s is an average speed of 10 m/s. That number describes the whole trip, though, not any single moment of it.

Picture a car that starts at a red light and speeds up steadily. Say it covers 20 m in 2 s. At the very start it was barely moving, and at the end it was moving faster than 10 m/s. Its speed kept changing, so the car had many different speeds during those 2 seconds.

Here is a question to think about: if the car was slow at the start and fast at the end, what would its speed be at the exact moment one second in, compared with the 10 m/s average?

You have two ideas in your answer, and they pull in different directions. Let's test the second one first.

Imagine the car's speed is a ramp: it starts at 0 m/s and climbs steadily to 20 m/s at the 2-second mark. Its average over those 2 seconds is 10 m/s, which matches your 20 m in 2 s.

Now picture a second car that runs at a steady 10 m/s for the whole trip. Its speed is 10 m/s at every moment, so the average and the instant speed are the same.

Here is the question: if the first car was going 0 m/s at the start and 20 m/s at the end, can its speed at the 1-second mark be 10 m/s? Tell me what you think it would show on a speedometer at that moment.

You reasoned that the speed climbs steadily from 0 to 20 m/s, so halfway through the 2 seconds it reads 10 m/s. That step is sound. Your conclusion that the average is the speed at 1 second is where I'd push back, though. The two numbers match here only because this particular speed-up is perfectly even. Let's test that idea on a different case.

Picture a car that waits at a red light for 1 second, sitting still at 0 m/s. Then it jumps to 20 m/s and holds that speed for the second second. Over the full 2 seconds it still covers 20 m, so its average is still 10 m/s.

Here is the question: at the 1-second mark, what does this car's speedometer read?

Yes, the speedometer reads 0 at exactly 1 second. The car is still waiting at the light, so the average of 10 m/s tells you nothing about that moment.

Now the instant question. Think about a short window of time right after the moment you care about. Suppose you want the speed at 1 second, and you look at how far the car travels between 1 second and 1.1 seconds. The time for that window is 0.1 s. If the car covers 2 m in that window, what is its average speed over those 0.1 s?

Use the same method as before: distance divided by time.

Your division is right: 2 m ÷ 0.1 s = 20 m/s. That is still an average, but over a much shorter window.

Think about what the car is doing in that window. It sits still until 1 s, then jumps to 20 m/s. So between 1 s and 1.1 s it is already moving at 20 m/s, and the average over that window is 20 m/s too.

Now shrink the window further. Suppose you look at the distance covered between 1 s and 1.01 s, which is 0.01 s long. If the car moves 0.2 m in that window, what is its average speed over those 0.01 s?

Your division is right: 0.2 m ÷ 0.01 s = 20 m/s, and the windows of 0.1 s and 0.01 s both give 20 m/s. You have also named the key idea: the window shrinks toward nothing, and the average approaches the speed at that instant.

Plugging in smaller numbers is a good start, and you can do it with a formula instead. Call the window length $\Delta t$ (the Greek letter delta means "a small change in"). If the car's position at time $t$ is $x(t)$, the average speed over the window from $t$ to $t + \Delta t$ is:

$$\frac{x(t + \Delta t) - x(t)}{\Delta t}$$

Think about the car from the red light again. Its position is 0 m until 1 s, then it moves at 20 m/s. Suppose you pick a window that starts at 1 s and ends at $1 + \Delta t$ seconds. What is the car's average speed over that window if $\Delta t$ is any positive number, such as 0.5 s? Tell me what you get, and whether the answer depends on the window length.