Suppose we have an element $x\in B$. Then it's image in $B_{f_i}$ is equal to some $F^i( \frac{b^i_1}{f_i^{k^i_1}} ,\ldots ,\frac{b^i_{j_i}}{f_i^{k^i_{j_i}}})$, where $\frac{b^i_j}{f_i^{k^i_j}}$ are the finite set of generators of $B_{f_i}$ over $A$, with $b^i_j\in A$, and $F^i$ some polynomials with coefficients in $A$. After multiplying by a large power of $f_i$ this gives us $n$ equalities in $B$ looking like $f_i^{N}x=F'^i(b^i_1,\ldots,b^i_{j_i},f_i)$, again with coefficients in $A$. But as $f_i$ generate unit ideal in $B$, there is an expression of 1 in terms of $f_i$: $a_1f_1+\cdots+a_nf_n=1$, with $a_i\in B$. Exponentiate it to the nN-th power and multiply by $x$, and you'll get $x=G(f_1,\ldots,f_n,a_1,\ldots,a_n,\ldots b^i_j \ldots)$, with $G$ polynomial with coefficients in $A$ (because after exponentiation each monomial of $a_i,f_i$ include at least one $f_j$ in power greater or equal to $N$, so after multiplying by x we could substitute $F'^j$ for $xf_j^N$). So any $x\in B$ can be expressed as a polynomial of $a_i,f_i,b^i_j$ with coefficients in $A$, which means that $B$ is finitely-generated $A$-algebra.

Looking it up in EGA as Georges suggests is also a good idea, I just thought you might not be ready for that yet. But I'm afraid he's got the reference wrong, and EGAI(6.6.3) is about different things. Your morphism is by definition (EGAI(6.3.1)) of finite type, and what you need is the fact that finite type for affine schemes $Spec(B)\to Spec(A)$ implies (actually, equivalent to) that $B$ is a finitely generated $A$-algebra, which is EGAI(6.3.3) on page 145. Though I may assure you that Maestro's proof is also long and technical.