Let $N, M$ be a linear spaces over the same field $\mathbb{F}$. So $\{e_i\}$, $\{\overline{e_i}\}$ -- two bases in $N$ and $\{e_k'\}$, $\{\overline{e_k}'\}$ -- two bases in $M$. Let $f$ be a map between $N$ and $M$, $f: N \rightarrow M$, and between bases $\{e_i\}$ and $\{e_k'\}$ it will be $A_f$ and between bases $\{\overline{e}_i\}$ and $\{\overline{e}_k'\}$ it will be $\overline{A_f}$. Let $B$ be the basis change matrix $B(\{e_i\}) = \{\overline{e_i}\}$ and similar for $C$, $C(\{e_k'\}) = \{\overline{e_k}'\}$.
So in my textbook I have this proof of the identity $\overline{A_f} = C^{-1} A_f B$:
$(\overline{e}_i)\overline{A_f} = f((\overline{e}_i)) = f((e_i)B) = (f(e_i))B = (e_i')A_f B = (\overline{e_i}')C^{-1}A_f B$
But I do not really understand why and how we can make follow steps: $\cdots = f((e_i)B) = (f(e_i))B = (e_i')A_f B = \cdots$. Why we can take the matrix $B$ beyond the parentheses? And why $f((e_i)) = (e_i')A_f$ and not $f((e_i)) = (e_i)A_f$?