I don't see what an Isomorphism does. It requires a bijective homomorphism: $$f(xy)=f(x)f(y)$$ What does this mean to me? When does this happen?
Also: An Endomorphism is a homomorphism mapping from a set to the same set, when would this ever not be bijective(e.g. an automorphism)?
Let's consider two groups, which I will call $G_1$ and $G_2$:
$$\begin{array}{c|c} G_1 & G_2\\ { \begin{array}{c|ccc} \star & 0 & 1 & 2 \\ \hline 0 & 0 & 1 & 2 \\ 1 & 1 & 2 & 0 \\ 2 & 2 & 0 & 1 \end{array}} & {\begin{array}{c|ccc} \ast & e & r & l \\ \hline e & e & r & l \\ r & r & l & e \\ l & l & e & r \end{array}} \end{array} $$
You can see from looking at the tables that there is a sense in which $G_1$ and $G_2$ are the same group: the two tables are exactly the same if you replace $0$ with $e$, replace $1$ with $r$, and replace $2$ with $l$.
In group theory we don't care what the group elements are called or how they are written, so $G_1$ and $G_2$ are the same groups, just written differently. This is what it means to be isomorphic. $G_1$ and $G_2$ are isomorphic groups.
How can we state mathematically that $G_1$ and $G_2$ are the same group, just with the elements given different names? We need to match up the elements of $G_1$ with the elements of $G_2$ so that elements that are the same in the two groups are matched up. So we have a function, say $f$, that tells us, for each element of $G_1$, which element of $G_2$ it is matched with. In this case we have $$0 \leftrightarrow e \\ 1 \leftrightarrow r \\ 2 \leftrightarrow l \\ $$
The function $f$ should be bijective, because we want each element of $G_1$ matched with exactly one element of $G_2$ and each element of $G_2$ matched with exactly one element of $G_1$.
Just having a bijective matching is not enough, because we can make such a matching between any two sets that are the same size. So we also want to show that the group operations of the two groups are the same. That is, whenever we have $a\star b = c$ in $G_1$, we also want to have $f(a) \ast f(b) = f(c)$ in $G_2$. For example, we have $1\star 2 = 0$ in $G_1$, so we should have $f(1) \ast f(2) = f(0)$ in $G_2$. With the $f$ from the previous paragraph, $f(1)\ast f(2) = f(0)$ says $r\ast l = e$, which agrees with the entry in the table for $G_2$.
Now this condition, that $$\text{for each $a,b,c$ in $G_1$,}\\ \text{if $a\star b = c$,}\\ \text{then $f(a)\ast f(b) = f(c)$}\\$$
is exactly equivalent to the condition that $f$ should be a homomorphism, because if $c = a\star b$ and $f(a)\ast f(b) = f(c)$ then $f(a) \ast f(b) = f(a\star b)$, just substituting the $c$ on the right-hand side of the equation. So we usually leave out $c$, replacing it with $a\star b$, to which it is equal, and then the condition looks like that one you were asking about. it still means that the $\star$ and $\ast$ operations are doing the same things in the two groups.
I hope this clears up something for you.
Two groups/rings/etc being isomorphic essentially means that two structures the same up to labelling. That is, given two algebraic structures $R$ and $S$, if $R\cong S$, then they're pretty much the same except for what you're calling the elements. For algebraic structures, the main focus isn't often on the elements, but on the operations; the isomorphism says that the operations act in the same regard, and serves as a sort of translator for the elements.
For example, $\mathbb{Z}\cong 2\mathbb{Z}$ as groups, which makes sense because we're really only changing what the generator is called.
As an example of an endomorphism $G\rightarrow G$ that isn't a bijection, just take the constant map sending every element in $G$ to the identity element of $G$, which isn't bijective if $G$ is non-trivial. A slightly less-trivial example is the map $\mathbb{Z}\rightarrow \mathbb{Z}$ mapping $n$ to $2n$.
Isomorphism is important because it allows one to tell when two algebraic objects are equivalent (that is, the only difference between them is the way in which the elements and operations are labelled). Two isomorphic objects have the same properties.
The definition of an endomorphism doesn't require bijection, an automorphism is a bijective endomorphism. That is to say, the set of automorphisms on a group (denoted $\operatorname{Aut}(G)$, usually), belongs to the set of endomorphisms of a group (denoted $\operatorname{End}(G)$, usually).
The relation can be expressed, automorphism $\implies$ endomorphism.
Automorphisms are subsets of Endomorphisms i.e. a linear map from a vector space to itself(In the case of vector spaces) and a group homomorphism from that group to itself. The difference is that Automorphisms are invertible whilst Endomorphisms aren't necessarily invertible.
