The symbol
$$n\choose k$$
is a shorthand for "The number of ways to choose $k$ objects from a collection of $n$ objects."
That is: you have $n$ objects in a line. You select $k$ of them, and paint them red, and you paint the other $n-k$ blue. How many ways are there to do that so that the final result looks different?
A moment of reflection will convince you that you could have chosen $n-k$ of the objects to paint blue first, and painted the rest of the objects red. The answer shouldn't depend on something arbitrary like whether you pick up the blue or the red paint first. If you can grok this fact, then you will have proved to yourself that
$${n\choose k} = {n\choose n-k}$$
How does this relate to binomial expansions? Well, say that we want to expand
$$(r + b)^n$$
Then we'll get a bunch of terms, each of which is a product of some $r$s with some $b$s. There are $n$ brackets, so there will be $n$ multipliers in each of the terms, and each of them will be either $r$ or $b$, so every term looks like
$$a_kr^kb^{n-k}$$
where $a_k$ is a coefficient, and $k=0, 1, 2, \dots n$.
What is $a_k$? Notice that in every bracket, we have to choose either an $r$ or a $b$. We can imagine going through the brackets one by one and making these choices, or you can imagine looking at all the brackets at once, and choosing which of them to pick $r$ form and which to pick $b$ from. The coefficient $a_k$ is the number of different ways of picking $k$ $r$s and $(n-k)$ $bs$ from the $n$ brackets. That is:
$$a_k = {n\choose k}$$
which, using the fact from earlier, we can instantly use to prove that the coefficients in a binomial expansion are symmetric:
$$a_k = {n\choose k} = {n\choose n-k} = a_{n-k}$$
Maybe you can see how binomial expansions relate to Pascal's triangle now?
Hint: It is linked to another cool and exciting relationship between the binomial coefficients!
$x - y = x+ (-y)= x+(-1)y$.
So $(x-y)^n =(x+(-y))^n=\sum\limits_{j=0}^n{n\choose j}x^{n-j}(-y)^j=$
$\sum\limits_{j=0}^n(-1)^j{n\choose j}x^{n-j}y^j$
It is presumed to be understood that $(-1)^j$ is $-1$ if $j$ is odd and is $1$ if $j$ is even.
So in essence: $(x- y)^n$ is much the same as $(x+y)^2$ but ever odd term is subtracted rather than added (and every even term is added).
So $(x+y)^n + (x-y)^n =\sum\limits_{j=0}^n{n\choose j}x^{n-j}y^j+ \sum\limits_{j=0}^n(-1)^j{n\choose j}x^{n-j}y^j=$
$\sum\limits_{j=0}(1+(-1)^j){n\choose j}x^{n-j}y^j=$
$\sum\limits_{j=0}^n\begin{cases}2 & \text{if j is even}\\0 &\text{if j is odd}\end{cases}{n\choose j}x^{n-j}y^j=$
$\sum\limits_{j=0;j \text{ is even}}^n2{n\choose j}x^{n-j}y^j$
$=$ "twice the even terms"
....
And $(x+y)^n - (x-y)^n=\sum\limits_{j=0}^n{n\choose j}x^{n-j}y^j- \sum\limits_{j=0}^n(-1)^j{n\choose j}x^{n-j}y^j=$
$\sum\limits_{j=0}(1-(-1)^j){n\choose j}x^{n-j}y^j=$
$\sum\limits_{j=0}^n\begin{cases}0 & \text{if j is even}\\2 &\text{if j is odd}\end{cases}{n\choose j}x^{n-j}y^j=$
$\sum\limits_{j=0;j \text{ is odd}}^n2{n\choose j}x^{n-j}y^j$
$=$ "twice the odd terms"
Best Answer
Since the size of the problem is small, we can count the cases directly. First, there are two partitions of 5 into at most 4 parts with each part at most 2, namely, $5=2+2+1$ and $5=2+1+1+1$. These correspond to the ways you can get $x^5$.
The partition $5=2+2+1$ means you get 2 factors of $-x^2$ from two of the terms, and a factor of $2x$ from another one. You can do this in $\binom42\binom21=12$. So you $12$ terms of the form $(-x^2)^2 2x$; in total $24x^5$.
For $5=2+1+1+1$ there are $\binom41\binom33=4$ cases and the terms are of the form $-x^2(2x) (2x)(2x)=-8x^5$; in total $-32x^5$.
So the $x^5$ term is $$ 24x^5-32x^5= -8x^5. $$
In your method you also need to expand $(2x-x^2)$ with a binomial sum.
Note: Corrections made based on the comment below.