From Jech's Set Theory:
We shall now define addition, multiplication and exponentiation of ordinal numbers, using Transfinite Recursion.
Definition 2.18 (Addition). For all ordinal numbers $\alpha$
- $\alpha + 0 = \alpha$,
- $\alpha + (\beta + 1) = (\alpha + \beta) + 1$, for all $\beta$,
- $\alpha + \beta = \lim_{\xi\to\beta}(\alpha+\xi)$, for all limit ordinals $\beta$.
Though he omits the details, which I'm trying to bring to light. After reading this post, it seems to me (please correct me if I'm wrong) that the sketch (some details are missing, hence the '$?$' signs) of the definition is as follows:
Definition (Addition). Fix $\alpha$. Define a function $G$ on $U=\{x:x=x\}$ such that, for any $\beta$-sequence $\{a_\xi:\xi<\beta\}$ we get $$ G : \{a_\xi\}\mapsto \begin{cases} \alpha & \text{ if $\beta=0$}\\ ? & \text{ if $\text{Suc}(\beta)$}\\ ? & \text{ if $\text{Lim}(\beta)$} \end{cases} $$ and, by transfinite recursion, there is a function $$F:\text{ON}\to U : \beta \mapsto G(F\upharpoonright \beta) = G(F_\xi : \xi<\beta) = \begin{cases} \alpha & \text{ if $\beta=0$}\\ F(\beta-1)+1 & \text{ if $\text{Suc}(\beta)$}\\ \lim_{\xi\to\beta}F(\xi) & \text{ if $\text{Lim}(\beta)$} \end{cases} $$
What is the meaning of $\sup\text{ran}(X)$ in the post? I do not recall seeing this terminology in the book. As $X$ is a transfinite (function) sequence $(a_\xi:\xi<\beta)$ it's range ($\text{ran}$?) is the collection of the $a_\xi$. I'm uncertain what the supremum ($\sup$?) of this class is though. If it is $a_\beta$, then $$ G : \{a_\xi\}\mapsto \begin{cases} \alpha & \text{ if $\beta=0$}\\ a_\beta+1 & \text{ if $\text{Suc}(\beta)$}\\ a_\beta & \text{ if $\text{Lim}(\beta)$} \end{cases} $$ but I do not know how the existence of $F$ follows from this $G$ by transfinite recursion. Furthermore, can we even add $1$ to $a_\beta$?
What exactly is $G$, and how does the existence of $F$ follow from that of $G$ by transfinite recursion?
Your comprehension is quite close. In your linked post, $\sup$ and $\operatorname{ran}$ indeed refer to supremum of a set and range of function respectively. The formal definition of supremum of a set of ordinals is just the union of them. The range of a set can be acquired by the replacement scheme.
So how do we add $1$ to $a_\beta$? In fact, $\gamma=\alpha+1$ is exactly another way to say $\gamma$ is the ordinal successor of $\alpha$, namely $\gamma=\operatorname{Succ}(\alpha):=\alpha\cup\{\alpha\}$. Thus you don't have to define general addition of ordinals in advance.
I would prefer to write $G(\langle a_\xi\rangle_{\xi<\beta})=\sup_{\xi<\beta}(a_\xi+1) = \bigcup(\operatorname{Succ}"\langle a_\xi\rangle_{\xi<\beta})$ since this covers both successor and limit cases, but in the following I will stick to the original definition in that post.
For the following $G$, $$ G(\varnothing) = \alpha \\ G(x)=\alpha \text{ if } x\text{ is not a function of ordinals} \\ G(\langle a_\xi\rangle_{\xi<\beta}) = a_{\beta_0}+1 \text{ if } \beta=\beta_0+1 \\ G(\langle a_\xi\rangle_{\xi<\beta}) = \cup_{\xi<\beta} a_\xi \text{ if } \beta \text{ is a limit ordinal} $$ the corresponding $F$ defined by Theorem 2.15 satisfies $F(\beta)=G(F\restriction\beta)$ for all ordinals $\beta$. From definition of $G$ we can tell that $$ F(\beta) = G(\langle F(\xi)\rangle_{\xi<\beta}) = \begin{cases} \alpha & \text{ if $\beta=0$}\\ F(\beta_0)+1 & \text{ if $\beta=\beta_0+1$}\\ \sup_{\xi<\beta} F(\beta_\xi) & \text{ if $\beta=\lim_{\xi<\beta}$} \end{cases} $$ as required.