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.