There are three issues here: well-ordering, ordinals, and transfinite induction.
Theorem. The following are equivalent in ZF:
(i) The Axiom of Choice: Given a nonempty family of nonempty sets $\{A_i\}_{i\in I}$, there exists a choice function for the family: a function $f\colon I\to\cup A_i$ such that $f(i)\in A_i$ for each $i\in I$.
(ii) Zorn's Lemma: If $(X,\preceq)$ is a partially ordered set in which every chain has an upper bound, then $X$ has $\preceq$-maximal elements.
(ii) The Well-ordering Principle: Given any set $X$, there exists an ordering $\leq\subseteq X\times X$ that is a well-ordering on $X$; that is, a total order on $X$ such that every nonempty subset of $X$ has a first element under $\leq$.
You can see a nice proof of the equivalence in this handout by George Bergman. (It's in Postscript).
So, assuming the Axiom of Choice, every set can be given a well-ordering.
Ordinals are special sets that generalize the natural numbers. One can prove that given two well-ordered sets $W_1$ and $W_2$, either $W_1$ and $W_2$ are order-isomorphic (there is a bijection between them that respects the order); or $W_1$ is isomorphic to an initial segment of $W_2$ (a subset $I$ such that if $a\in I$ and $b\in W_2$ with $b\leq a$, then $b\in I$); or $W_2$ is isomorphic to an initial segment of $W_1$.
The idea is to define ordinals to be sets where we define an ordering $\alpha\lt\beta$ if and only if $\alpha\in \beta$, and where each ordinal $\alpha$ is given by $\alpha=\{\beta\mid\beta\text{ is an ordinal, and }\beta\in\alpha\}$.
A set $T$ is transitive if $x\in T\Rightarrow x\subseteq T$. We define
Definition. An ordinal is a transitive set that is well-ordered by $\in$.
Under this definition: $0=\emptyset$ is an ordinal; if $\alpha$ is an ordinal and $\beta\in\alpha$, then $\beta$ is an ordinal; if $\alpha$ and $\beta$ are ordinals, and $\alpha\subsetneq \beta$, then $\alpha\in\beta$; and if $\alpha$ and $\beta$ are ordinals, then either $\alpha\subseteq \beta$ or $\beta\subseteq \alpha$. If $\alpha$ is an ordinal, then so is $\alpha^+ = \alpha\cup\{\alpha\}$.
Also, every well-ordered set is order isomorphic to one and only one ordinal.
The finite ordinals are precisely the natural numbers. But there are other ordinals.
An ordinal $\alpha$ such that $\alpha=\beta^+$ for some ordinal $\beta$ is called a "successor ordinal". If $\alpha$ is not a successor ordinal, then $\alpha=\sup\{\beta\mid \beta\lt \alpha\}=\cup\alpha$, and $\alpha$ is called a "limit ordinal."
The finite ordinals correspond to the natural numbers; you can think of them as just finite ordered lists. So, for example, the ordinal $5$ can be thought of as a list with five items, namely the elements $0$, $1$, $2$, $3$ and $4$, in that order: $5:\qquad \stackrel{0}{\bullet}\quad \stackrel{1}{\bullet}\quad \stackrel{2}{\bullet} \quad \stackrel{3}{\bullet}\quad\stackrel{4}{\bullet}$ The first infinite ordinal is $\omega$, which corresponds to the set of all natural numbers in their usual order. $\omega: \qquad \stackrel{0}{\bullet}\quad \stackrel{1}{\bullet}\quad \stackrel{2}{\bullet} \quad \stackrel{3}{\bullet}\quad \cdots \quad\longrightarrow$ where $\longrightarrow$ is meant to represent that it keeps going.
The next ordinal after $\omega$ is $\omega^+$; it is like having the natural numbers, and then another item which is strictly larger than every natural number: $\omega+1: \qquad \stackrel{0}{\bullet}\quad \stackrel{1}{\bullet}\quad \stackrel{2}{\bullet} \quad \stackrel{3}{\bullet}\quad \cdots \quad\longrightarrow\quad \stackrel{\omega+1}{\bullet}$
Then $(\omega^+)^+$ (usually written $\omega+2$), which is a copy of the natural numbers, and then two more items larger than all natural numbers: $\omega+2: \qquad \stackrel{0}{\bullet}\quad \stackrel{1}{\bullet}\quad \stackrel{2}{\bullet} \quad \stackrel{3}{\bullet}\quad \cdots \quad\longrightarrow\quad \stackrel{\omega+1}{\bullet}\quad\stackrel{\omega+2}{\bullet}$
And so on. Later, you get to $\omega+\omega$, two copies of the natural numbers (say, blue and red natural numbers, with every blue natural number smaller than every red natural number): $\omega+2: \qquad \stackrel{0}{\bullet}\quad \stackrel{1}{\bullet}\quad \stackrel{2}{\bullet} \quad \stackrel{3}{\bullet}\quad \cdots \quad\longrightarrow\quad \stackrel{\omega+1}{\bullet}\quad\stackrel{\omega+2}{\bullet}\quad\stackrel{\omega+3}{\bullet}\quad\cdots\quad\longrightarrow$ And so on.
Finally, transfinite induction:
Transfinite Induction. Let $C$ be a class of ordinals, and assume that:
- (i) $0\in C$;
- (ii) If $\alpha\in C$, then $\alpha^+\in C$.
- (iii) If $\alpha$ is a nonzero limit ordinal, and $\beta\in C$ for all $\beta\in \alpha$, then $\alpha\in C$.
Then $C$ is the class of all ordinals.
This generalizes usual induction (which would only use parts (i) and (ii)). The main difference is how to deal with "limit ordinals"; the process is similar to that used in strong induction: assume valid for every ordinal strictly smaller, prove it for the current ordinal.
So now, looking at the proof offered by Robinson. Your group $F$ has a basis $X$, because it is free abelian. Invoking the Well Ordering Principle, we may assume that $X$ is actually well-ordered; since every well-ordered set is isomorphic to an ordinal, we may go so far as to just assume that that the basis is given as a set indexed by an ordinal $\beta$, $X=\{x_{\alpha}\}_{\alpha\in\beta}$.
For each ordinal $\alpha$, let $F_{\alpha}=\langle x_{\gamma}\mid \gamma\lt\alpha\rangle$. So $F_{\alpha}=F$ for any $\alpha\geq\beta$. We are going to prove by transfinite induction that the class of all ordinals $\alpha$ such that
$H_{\alpha} = H\cap F_{\alpha}$ is free abelian of rank at most $\alpha$.
is the class of all ordinals. Since $H_{\beta}=H$, if we can prove that the class is the class of all ordinals, we will have our desired result (that $H$ is free).
So we proceed by transfinite induction. $F_{0}=\{0\}$, and $H\cap F_{0}=\{0\}$ is free on the empty set, so we are done. $0$ has the property.
Assume that $\alpha$ has the property that $H_{\alpha}$ is free abelian, and consider $H_{\alpha+1}$. If $\alpha\geq \beta$, there is nothing to do: $H_{\alpha+1}=H_{\alpha}$, hence it is free abelian, so we may assume that $\alpha\lt\beta$. Then proceeding as Robinson does, we see that $H_{\alpha+1}$ is either equal to $H_{\alpha}$, or is isomorphic to $H_{\alpha}\oplus \mathbb{Z}$, and since $H_{\alpha}$ is free abelian, so is $H_{\alpha}\oplus\mathbb{Z}$, hence $H_{\alpha+1}$ is free abelian. Thus, if $\alpha$ has the property, then so does $\alpha+1$.
Finally, we need to show the result holds if $\alpha$ is a limit ordinal; Robinson is omitting this part (I'm not sure why; maybe he has a different proof in mind). We know that for every $\gamma\lt \alpha$, $H_{\gamma}$ is free. It is not hard to see that $F_{\alpha} = \bigcup_{\gamma\lt\alpha}F_{\gamma}$ and so $H_{\alpha} = H\cap F_{\alpha} = H\cap\left(\bigcup_{\gamma\lt\alpha}F_{\gamma}\right) = \bigcup_{\gamma\lt\alpha}H\cap F_{\gamma} = \bigcup_{\gamma\lt\alpha}H_{\gamma}.$ And it is, in turn, not hard to verify that this union is free abelian (you need to be careful, but you can pick bases for $H_{\gamma}$ for each $\gamma$, again inductively, in such a way that if \gamma'\lt\gamma, then the basis of H_{\gamma'} is a subset of the basis of $H_{\gamma}$; then the union of the bases of the $H_{\gamma}$ is a basis for $H_{\alpha}$); and that the cardinality of this basis is at must the supremum of the cardinals of the bases of the $H_{\gamma}$, which is bounded above by $\alpha$ by the inductive hypothesis.
So the conclusion also holds for limit ordinals, hence holds for all ordinals (since it holds for $0$, for successor ordinals, and for limit ordinals).