As $A$ is a finite algebra over $K$ it is noetherian. As $M$ is finitely generated there is a surjection $A^{\oplus n} \longrightarrow M$. $A^{\oplus n}$ is noetherian as $A$ is. Let $N$ be the kernel of this map. By noetherianness, it is finitely generated, so there is a surjection $A^{\oplus m} \longrightarrow N$ and hence an exact sequence $A^{\oplus m} \longrightarrow A^{\oplus n} \longrightarrow M \longrightarrow 0$. Repeat this process to get a projective (in fact free) resolution by finitely generated modules. Note that we didn't need the full strength of the assumption that $A$ is finite over $K$ - only that it was noetherian.
If $M$ is projective, then there is module $K$ s.t. $M \oplus K \cong R^{(\Lambda)}$. For any multiplicative set $S \subset R$, one gets $S^{-1}M \oplus S^{-1} K \cong (S^{-1}R)^{(\Lambda)}$. Hence, any localization of a projective module is projective.
If $(R,m)$ is local and $M$ is finitely generated projective then $M$ is free. Here is a proof. The short exact sequence
$$
0 \to K \to R^n \to M \to 0,
$$
splits, hence $K$ is finitely generated and sequence
$$
0 \to K \otimes R/m \to R^n \otimes R/m \to M \otimes R/m \to 0,
$$
also splits. By Nakayama's lemma we can choose generators of $M$ in such way that the map $R^n \otimes R/m \to M \otimes R/m$ is an isomorphism, then $K \otimes R/m \cong K/mK=0$. By Nakayma's lemma $K=0$, since it is finitely generated.
Fact: if $M$ is finitely presented, then $S^{-1}Hom_R(M,N) \cong Hom_{S^{-1}R} (S^{-1}M, S^{-1}N)$.
To check that $M$ is projective is enough to check that for any surjection $N \to N'$ map $Hom(M,N) \to Hom(M, N')$ is surjective, but this can be checked stalkwise if $M$ is finitely presented (by the fact).
There are plenty of other proofs, but often they involve flat modules, in this proof we only use two ways to characterize projective modules and the fact. Notice that this proof doesn't use that $R$ is Noetherian.
Best Answer
A regular local ring $R$ is Noetherian by hypothesis. If $M$ is a finitely generated $R$-module, then there is a finite rank free module $F$ and a surjection $F\to M$ fitting into a short exact sequence $$0\to N\to F\to M\to0.$$ As $N$ is a submodule of the finitely generated free module $F$ over the Noetherian ring $R$ then $N$ is finitely generated. Iterating this, gives a resolution of $N$ by finitely generated free modules. As $R$ is regular, it has finite global dimension, and the iterated kernels are eventually projective (so free) and the resolution can be brought to an end.