Let's forget about numbers for a minute and consider something easier. You know that people have mothers and some people are the mothers of other people.
Then $$\forall b\exists a. \text{person $a$ is the mother of person $b$}$$ means that for each person $b$, there is some person $a$, such a that $a$ is the mother of $b$. In other words, each person $b$ has a mother. This is true.
But $$\exists a \forall b. \text{person $a$ is the mother of person $b$}$$ says that there is a person $a$ such that for each person $b$, $a$ is the mother of $b$. Here there is one person $a$ who is the mother of everyone. This is false. (Not as a matter of logic, but as a matter of fact: there isn't anyone who is the mother of everyone. If you think there might be, who do you suppose it is? Angela Merkel perhaps? No, she's not the mother of everyone because she's not my mother Okay, perhaps my mother then? No, my mother is not the mother of everyone either; she's only the mother of me and my sister. No, there really isn't anyone who is everyone's mother.)
Now let's go back to numbers.
$$\forall b\exists a. a>b$$
says that for each number $b$, there is some number $a$ that is bigger. This is true, because If someone gives you some number $b$ and dares you to name a number $a$ that is bigger, you can just reply with $a=b+1$, which is bigger than $b$.
But this
$$\exists a\forall b. a>b$$
is false, because it says that there is some number $a$ that is bigger than every number $b$. And of course there isn't one; if you think you have an example, say $a$, I can point out that it is not bigger than $a+1$.
Sometimes it's helpful to think of the quantifiers as a game. The $\exists$ are your moves in the game, and the $\forall$ are your opponent's moves. To show that something like $$\forall b\exists a. a>b$$ is true, your opponent selects a $b$, and then you have to select an $a$ that makes $a>b$ true. You can always do this, by picking $a = b+1$.
But to show that $$\exists a\forall b. a>b$$ is true, you have to go first: you pick an $a$, and then your opponent gets to pick $b$, and you win only if $a>b$. But your opponent can defeat you you by picking a $b$ that is smaller than $a$.
Sometimes it does happen that you can win a game like this one. Suppose that instead of $a>b$ we had $a\cdot b=0$. Then you're playing the game $$\exists a\forall b. (a\cdot b=0).$$ To be sure to win, you need to pick an $a$ so that no matter what $b$ your opponent picks on his move, you get $a\cdot b=0$. There is only one winning move here. Can you find it?
