What is the definition of a free abelian group generated by the set $X$?

The free abelian group $G(X)$ generated by $X = \{A,B,C,D\}$ is the smallest abelian group containing $X$ and has no other restricting properties. You should think of $X$ as a basis for the group $G(X)$, similarly to a basis for a vector space.

Concretely, what are the elements of $G(X)$? Starting from $X$, first adjoin symbols that will represent the identity in $G(X)$ and inverses for the generating elements: $0, −A, -B, -C, -D$. Now form the set of all abstract finite sums, e.g. $A + B + D + (-B)$. Finally, introduce relations (i.e. define the set of equivalence classes of sums) saying that these elements and sums behave like a commutative group; for example, $A + B + D + (-B) = A + D$.

Since the group is abelian, we can collect terms; for instance, $A + B + A = A + A + B = 2A + B$. So, each element of the free abelian group $G(X)$ can be expressed uniquely as a sum $$ n_AA + n_BB + n_CC + n_DD $$ where each $n_x \in \mathbb{Z}$; in other words, $G(X) = \big\{\sum_{x \in X} n_X X \mid n_x \in \mathbb{Z} \big\}$. Therefore, you can easily see that $G(X) \simeq \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}$, where the group operation is the pointwise addition and the neutral element is $(0,0,0,0)$.

In the explanation above, note that actually the nature of the elements of $X$ plays no role, the only relevant information is the number of elements of $X$. Thus, more generally, you can see that given a set $X$ with exactly $n \in \mathbb{N}$ elements, $G(X) \simeq \underbrace{\mathbb{Z} \times \dots \times \mathbb{Z}}_{n \text{ times } \mathbb{Z}}$.

It is the abelian group $ \mathbf Z^{(X)}$ i.e. the set of functions $f:X\rightarrow \mathbf Z$ with finite support. If, as is usual, we represent such a function by its values at each point $x\in X$, it is $$ \mathbf Z^{(X)}=\Bigl\{(n_x)_{x\in X}\mid n_x=0\enspace\text{except for a finite number of them}\Bigr\}.$$

If $|X|=n$, it'is the same as $\mathbf Z^X=\displaystyle\prod_{x\in X}\mathbf Z$ and it's isomorphic to $\mathbf Z^n$.

Note: as an abelian group, the ring of polynomials with integer coefficients is defined as $\mathbf Z^{(\mathbf N)}$, to which one adds an extra multiplicative structure. In contrast, $\mathbf Z^{\mathbf N}$ with the extra multiplicative structure given by the Cauchy product, is the ring of formal power series with integer coefficients.

One way you can think of this in a very elementary and canonical way is that the free abelian group on a finite set such as $\{A,B,C,D\}$ is the group of all functions $$f : \{A,B,C,D\} \to \mathbb{Z} $$ with the operations of pointwise addition and pointwise inverse: given any two such functions, for all $x \in \{A,B,C,D\}$ define $$(f_1+f_2)(x) = f_1(x) + f_2(x) $$ also $$(-f)(x) = -f(x) $$ The identity element, of course, is the constant function $f(x)=0$.

For infinite sets, you have to add one more clause to the definition: for the free abelian group on a set $X$, you do not take all functions $f : X \to \mathbb{Z}$, you take only those functions which satisfy the property that there is a finite subset $Y \subset X$ such that if $x \in X-Y$ then $f(x)=0$.