All rings are unital commutative. Let $\pi:B \rightarrow A$ be a ring homomoprhism. Let the image of $f \in B$ be $f'$. Then $A_{f'}$ is a $B_f$-algebra. It seems to me that the following is true.
$A_{f'}$ is f.g. as a $B_f$-algebra iff it is f.g. as a $B$-algebra.
My argument: Suppose it is f.g. as $B_f$-algebra. Let the generators be $a_i/(f')^{k_i}$. The way $B_f$ acts on an element $A_{f'}$ is via
$$ \frac{b}{f^k} \cdot \frac{a}{{f'}^l} = \frac{\pi(b) \cdot a}{{f'}^{k+l}} .$$
In other words $A_{f'}$ is in fact generated by $a_i, \frac{1}{f'}$ as both $B$-algebra.
Equally, if it is f.g. as a $B$-algebra, then the same argument shows it is f.g. as a $B_f$-algebra.
I would be happy to know if the claim is true and if my argument is correct, if not, what went wrong.
Your argument looks good to me. Here is another take: $B_{f} \cong B[X]/\langle fX - 1 \rangle$, and the data of a $B_{f}$-algebra $A$ is equivalent to the data of $B$-algebra $A$ such that the image of $f$ under the structural map $B \to A$ is invertible - this follows from the universal property of localization.
Now, a $B$-algebra $A$ is finitely generated iff it admits a surjective $B$-algebra homomorphism $\varphi \colon B[X_{1}, \ldots, X_{n}] \to A$. If the image of $f \in B$ in $A$ is invertible, then by the universal property of localization, $\varphi$ extends to a (surjective) $B_{f}$-algebra morphism $\rho \colon B_{f}[X_{1}, \ldots, X_{n}] \to A$.
On the other hand, suppose that $A$ is a finitely generated $B_{f}$-algebra, i.e. $A$ admits a surjective $B_{f}$-algebra morphism $\rho \colon B_{f}[X_{1}, \ldots, X_{n}] \to A$. Since $B_{f}[X_{1}, \ldots, X_{n}] \cong B[X_{0}, X_{1}, \ldots, X_{n}]/\langle fX_{0} - 1 \rangle$ as $B$-algebras, the composition of $\rho$ with the natural surjection of $B$-algebras $B[X_{0}, X_{1}, \ldots, X_{n}] \to B_{f}[X_{1}, \ldots, X_{n}]$ gives a surjective $B$-algebra map $\varphi \colon B[X_{0}, X_{1}, \ldots, X_{n}] \to A$.