Understanding the Definition of Well-Founded Induction

Let $A$ have a well-founded order $\prec$. A proof by well-founded induction would look like this:

Theorem. For all $x\in A$ we have $P(x)$.

Proof. Let $a\in A$ be arbitrary. Assume that $P(b)$ holds for all $b$ with $b\prec a$. Then (bla bla bla ... some argument ... bla bla). Therefore, also $P(a)$. Thus the claim of the theorem follows by well-founded induction. $\square$

You may note that this proof skeleton is very similar to what is called "strong induction" for the natural numbers. And as to why well-founded induction works is indeed very similar to why (strong) induction works on the natural: The central point of the proof shows

If $P(b)$ holds true for all $b\prec a$, then $P(a)$

which is the same as saying

If $\neg P(a)$, then there exists $b\prec a$ with $\neg P(b)$

or succinctly

There is no smallest counterexample

And this the shows that there is no counterexample at all because for any counterexample $a_0$, the continued picking of smaller counterexamples $a_1\prec a_0$, $a_2\prec a_1$, etc. would give us an infinite descending chain.

I think the key to resolving your confusion lies in this paragraph:

Finally, the "second big block" is saying if $y$ precedes $x$, then $P(y)$ is true. So if $y$ comes before $x$, then we at least know $P(y)$ is true. Not really sure what this means (how we know $P(y)$ is true).

Note that we don't have to know how $P(y)$ is true ... we just have to show that if $P(y)$ is true for all $y$ that are 'smaller' than $x$, then $P(x)$ will hold as well. For if we can show that, then indeed all objects will have property $P$.

And why is that?

Well, take any arbitrary object of the set. Given that there is no infinitely descending chain, there are only two possibilities:

A. The object has no 'smaller' element. We call such an element a 'minimal' element. There may be any number of 'minimal elements, but every minimal element $x$ will have to have the property $P$ once we were able to show that if $P(y)$ is true for all $y$ that are 'smaller' than $x$, then $P(x)$ will hold as well. And that is because with no smaller elements, it is vacuously true that $P(y)$ is true for all $y$ that are 'smaller' than such a minmal element, and hence the minimal element will have to have the property $P$. As such, these 'minimal elements' are of course the base cases of the induction.

B. The object is some finite number of steps 'up' from the base cases. Well, given that those base cases all have the property, then their successors all have the property as well, etc. ... and given that any object is only some finite number of steps up from the base cases, eventually this step-by-step 'propagation' of objects having the property $P$ will reach the object under consideration as well.

In the end, the logic here is really very similar to the logic behind strong induction for the natural numbers. It's just generalized to point out that it will work for any well-founded set.