First: a Galois field is a finite field. A finite field will have $p^n$ elements for some prime $p$ and some positive integer $n$ (in fact, there is, up to isomorphism, one and only one finite field with $p^n$ elements). If a field has $p^n$ elements, then $x+x+\cdots+x=0$ ($p$ summands) for all $x$, and no smaller positive integer has that property; this is called the characteristic of the field. For fields of order a power of $2$, you do indeed have $x+x=0$ for all $x$, but not for any other size Galois fields. However, in computer science and cryptography there is some preference for working in fields of characteristic $2$, because they tend to be easier to represent and work with in computers (which themselves work in "characteristic $2$").
I'm not sure what you mean by "why do I start dividing $x^0$, $x^2$, $x^2$, etc.?". As you have probably seen, the field $GF(2^n)$ can always be described in terms of a monic polynomial $P(t)$ of degree $n$ that is irreducible over $GF(2)$. This amounts to constructing the field $GF(2)[t]/(P(t))$. What this tells you is that you have the smallest ring that contains $0$, $1$, and $x$, subject to the conditions $1+1=0$, $1\cdot 1=1$, and $P(x)=0$. Any element of this ring can be written as a polynomial expression in $x$ with coefficients $0$ or $1$, $q(x) = b_mx^m+\cdots + b_0$.
Now, because we are assuming that $P(x)=0$, using the division algorithm we can always write any polynomial $q(t)$ as $q(t)=P(t)b(t) + r(t)$, with $r(t)=0$ or $\deg(r)\lt \deg(P)=n$. Evaluating at $x$ and using the fact that $p(x)=0$, we get that $q(x)=r(x)$, so in fact every element in this ring can be written as a polynomial expression in $x$ of degree less than $n$. The expression is unique, because if $r(x)=s(x)$ with $r$ and $s$ of degree less than $n$, then $r-s$ would be a multiple of $P(t)$ (because the latter is irreducible), and degree considerations tell you that $r=s$.
So each element of $GF(2^n)$ can be written uniquely as $q(x)$ where $q$ has degree less than $n$. Choosing different polynomials amounts to choosing different representations for the elements, and so different rules for what $x^n$ will mean.
So: in order for $x^2=x+1$ to hold in your Galois field $GF(p^n)$, you must have that $x^2-x-1=0$, so the polynomial $P$ you picked must divide $x^2-x-1$, and so you must be either in $GF(p)$ or in $GF(p^2)$. Similarly, for $x^3=x^2+x=1$ to hold you would need to be in $GF(p)$ or $GF(p^2)$ in order for $x^2+x-1=0$ to hold; and whether it holds depends on the polynomial $P$ that you picked to define how $x$ behaves.
In summary: you divide by the polynomial so you can get a unique expression for each element in $GF(2^n)$ in terms of the distinguished element $x$ that helps you construct it (the behaviour of the element $x$ being determined by the polynomial). The condition $\alpha+\alpha=0$ will hold for every $\alpha$ in every Galois field of size $2^n$ for any $n$; but no power of $x$ smaller than $n$ will be expressible in terms of lower powers of $x$, and the expression of $x^n$ in terms of lower powers is determined by your choice of polynomial.