For the case $R$ is a local ring it's a corollary of Nakayama's lemma.
As the notation in the above link, suppose $M$ is a finite generated projective module over $R$, then, first pick a minimal number of generators, i.e., $M=Rm_1+\cdots +Rm_k$, and $k$ is the minimal number with this property, so we get a decomposition
$$R^k=M\oplus N,$$ then, we are left to prove $N=0$.
First, applying $R/I\otimes-$, where $I$ is the unique maximal ideal in $R$, then we get $$(R/I)^k=M/IM\oplus N/IN,$$ and note that $M/IM$, $N/IN$ are vector spaces over the field $R/I$, so by comparing the dimension, we get $N/IN=0$, i.e., $N=IN$, then,
we use the Nakayama's lemma, the Statement 1 in the above link, we get $r\in 1+I$, such that $rN=0$, but $r\not \in I$ and $R$ is local implies $r$ is a unit, so $N=0$.
Remarks. 1) To get the choice of $k$, we can first assume $k=\dim_{R/I}(M/IM)$, then use the Statement 4 in the above link to lift the basis of $M/IM$ to get a minimal set of generators of $M$.
2) A deep theorem of Kaplansky says that any projective modules (not necessarily finitely generated) over a local ring is free.
It is always true that $K(A)$ is finitely generated as a module over $K(B)$ .
Since you mention quotient fields, we may assume that $A$ is a domain.
We may also assume that $B\subset A$ and thus that $B$ is a domain too: if this is not the case and you have an algebra $\phi:B'\to A$, just consider $A$ as a $B=\phi(B')$-algebra.
With these preliminaries out of the way, let $S=B\setminus \lbrace 0\rbrace$.
Then (by base change or by hand) $S^{-1}A$ is a finitely generated $S^{-1}B$-module .
Since $S^{-1}B$ is a field and $S^{-1}A$ is a domain, $S^{-1}A$ is a field too by the Useful Lemma below.
Finally,we have $A\subset S^{-1}A\subset K(A)$ and since $S^{-1}A$ is already a field, we have $S^{-1}A=K(A)$ : we have proved that $K(A)=S^{-1}A$ is a finitely generated module (=finite-dimensional vector space ) over the field $K(B)=S^{-1}B$.
Useful Lemma:
If the commutative domain $R$ is a finite-dimensional algebra over a field $K$, then $R$ is a field.
Proof:
Let $0\neq r\in R$. The multiplication map $m_r:R\to R:x\mapsto rx$ is an injective endomorphism because $R$ is a domain, hence it is surjective by elementary linear algebra .
Thus there exists $s\in R$ with $m_r(s)=rs=1$ and thus $r$ is invertible: $r^{-1}=s$.
NB I have simplified my original proof by invoking the Useful Lemma instead of integral extensions.
Best Answer
If $I$ is an ideal of $R$, the dual of $R/I$ is isomorphic to $\mathrm{Ann}(I) = \{r \in R : rI = 0\}$, and this doesn't have to be finitely generated. Take for instance $R = k[y,x_1,x_2,\dotsc]/(y x_i : i \geq 1)$ and $I=(y)$.
A more natural question would be: How can we characterize commutative rings with the property that duals of f.g. modules over that ring are f.g.? Even more natural: How can we characterize commutative rings with the property that hom modules between f.g. modules are f.g.? For example, noetherian commutative rings satisfy this property (see the comment by Keenan Kidwell). But I think that there are more examples (perhaps coherent rings?).