You need this to do induction on $ON$. The induction you know from analysis is over $\omega \in ON$. The reason why you want induction over $ON$ is because you might want to prove something like this:
Let $X$ be a set, let $\mathcal{C}$ be a family of subsets of $X$ that is closed under unions of subfamilies that are linearly ordered by inclusion and let $f : \mathcal{C} \to \mathcal{C}$ be a function with $f(C) \supseteq C$ for all $C \in \mathcal{C}$. Then there exists a $C$ in $\mathcal{C}$ such that $f(C) = C$.
Here's an outline of the proof: You pick an ordinal $\alpha$ that cannot be mapped into $\mathcal{C}$ injectively. Then you define a function $F: \alpha \to \mathcal{P}(X)$ involving $f$ such that certain properties hold. The definition of $F$ is where the transfinite recursion comes in because $F$ needs to be defined for all $\alpha$ in $ON$. The definition looks something like this:
Assume $\beta \in \alpha$ and $F(\gamma)$ has been defined for all $\gamma < \beta$. Then
$ F(\beta) := \begin{cases} f(\bigcup\{ F(\gamma) : \gamma < \beta \}) ) & \text{if } \bigcup\{ F(\gamma): \gamma < \beta\} \in \mathcal{C} \\ X & \text{otherwise} \end{cases}$
You can find the full proof of this theorem on page 163 of "Discovering Modern Set Theory" by Just and Weese. If you don't like Kunen have a look at that. I have both and I think Kunen is too painful to read.
For another example of transfinite recursion see here. I'd recommend that you first look at an example of transfinite induction, like for example the proof of $\alpha + \beta = \alpha \cup \{ \alpha + \gamma : \gamma < \beta \}$ for ordinals $\alpha, \beta$. Then you'll get the idea of doing the induction in two parts, once for limit ordinals and once of "normal" ordinals. And from there it's easy to understand what transfinite recursion is -- they're almost the same thing.
Hope this helps.