I'm trying to fill in some gaps regarding the definition of ordinal arithmetic. In particular, we define $$ \alpha +_{ON} \beta = \begin{cases} \alpha & \beta = 0, \\ S(\alpha +_{ON} \gamma) & \text{if } \beta = S(\gamma), \\ \cup(\{\alpha +_{ON} \delta : \delta < \beta\}) & \text{if $\beta$ is a limit ordinal.} \end{cases} $$ By the transfinite recursion theorem, I understand that this is a sufficient definition of ordinal addition, since the function furnished by the theorem is unique. More specifically, for each $\alpha \in ON$ we take the class function $F: V \rightarrow V$ given by $F(x) = \alpha$ if $x$ is $0$, $F(x) = S(F(\beta))$ if $x$ is a function with domain the successor ordinal $S(\beta)$, $F(x) = \cup(\text{ran}(x))$ if x is a function with domain a limit ordinal, and $F(x) = 0$ otherwise. This $F$ produces the unique class function $g:ON\rightarrow V$ such that $g(\delta) = F(g \upharpoonright \delta)$. Then $g(\beta) = \alpha +_{ON} \beta$.
My question is, how can we show/know that $F$ is actually a class function, since $F$ itself is defined recursively (when the input is a function with successor ordinal as its domain)?
The successor clause in the definition of $F$ has a typo. It should be $F(x)=S(x(\beta))$ when $x$ is a function with domain $S(\beta)$. So the correct definition of $F$ is explicit, not recursive.