Subset axioms$\;\;\;$ For each formula $\varphi$ not containing $B$, the following is an axiom: $$\forall t_{1}\cdots\forall t_{k}\,\forall c\,\exists B\,\forall x(x\in B \iff x\in c \;\&\; \varphi)$$
Choice axiom$\;\;\;\;\;(\forall\; \text{relation}\;R)(\exists\; \text{function} \;F)(F \subseteq R \;\&\; \text{dom}\; F = \text{dom}\; R)$
If $\varphi$ can contain $B$, axiom of choice could be derived from subset axiom. But why can't $\varphi$ contain $B$?
Intuitively, because the point of the axiom is to define the set $B = \{x \in c \mid \varphi\}$. So we can't use $B$ in the definition of $B$.
Looking directly at the axiom, one thing that could go wrong if we allowed $\varphi$ to contain $B$ is the special case $$ \forall t_{1}\cdots\forall t_{k}\,\forall c\,\exists B\,\forall x(x\in B \iff x\in c \land x\notin B) $$ This can't possibly be the case for any $c \ne \varnothing$, because for every element of $c$, it says that $x \in B \iff x \notin B$.