Understanding of definitition of Ordinals Definition. An ordinal is a well-ordered set $\alpha$ such that for all $x\in\alpha$ 

$x\in$ $\left( -\infty ,x\right)$
in other words,
$y\in x\Leftrightarrow y \in \left( -\infty .x\right) \Leftrightarrow $y$ < $x
i.e., the set $\alpha$ is well-ordered by the element ship relation $\in$.
It follows from the defitinition that if $\alpha$ is an ordinal then every element of $\alpha$ is proper initial segment of $\alpha$.
My question is that why $x\in$ $\left( -\infty ,x\right)$, ,is there a problem?
 A: What you probably meant is $x = (-\infty,x)$. In other words "each element $x$ of the ordinal (remember, $x$ is also a set, since everything ever is a set) should be equal to the set of all smaller elements.
The reason for this is that, as you mentioned, it means that the two orders on $\alpha$ agree. See, every set you've ever seen in your life comes with it's own built-in partial order order, just because everything is a set, and we can ask when sets contain eachother. (Many sets don't have any inclusions among their elements at all, so this partial order is often trivial.) You called $\alpha$ a well-ordered set, so that means it comes with an order of its own. The condition that these two orders agree can be rewritten as
$$y<x\Leftrightarrow y\in x,$$
as you pointed out.
Some people follow the a different approach to defining ordinals: they just look at all well-ordered sets, and then say that if a copy of one $A$ sits inside of another $B$, then $A\le B$ and if they can sit inside of eachother (even if the two inclusions are totally unrelated) then they must be "they same" order. These equivalence classes are called ordinals, and two well-ordered sets are said to "have the same ordinality" if they're equivalent in this sense.
The approach you're being exposed to might be thought of as picking one special $\alpha$ in each equivalence class. There are even still several ways to do this, and your source chose to do so by saying "for each ordinal class, our special chosen well-ordered set $\alpha$ is the one whose order is the magical, totally-for-free order that pervades everything: containment of sets."
EDIT-ISH: I commented previously that the set needed to be transitive (i.e., contain all the elements of its elements). This is true, but vacuous: any set satisfying your condition is automatically transitive, since every element of $x\in\alpha$ automatically in another element of $\alpha$.
