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.
Yes, a surjective $A$-linear endomorphism $T:M\to M$ of a finitely generated $A$-module $M$ is an isomorphism.
Proof
Consider $M$ as an $A[X]$-module via the multiplication $P(X)*m=P(T)(m)$ , so that for example $(X^2-7)*m=T(T(m))-7m$.
[this is a classical trick used in advanced linear algebra].
Surjectivity of $T$ translates into $M=XM$ and so a fortiori for the ideal $I=XA[X]$ we have $M=IM$.
Now Nakayama's lemma comes to our rescue : it says that there exists an element $XQ(X)\in I$ such that $(1-XQ(X))*m=0$ for all $m\in M$, which means that $m=TQ(T)m$ and this immediately implies that $T$ is invertible with inverse $T^{-1}=Q(T)$.
This result and its extremely elegant proof are due to Vasconcelos.
And if, like so many of us , you keeep forgetting what Nakayama says, look here.
Caveat
Of course an injective endomorphism of a finitely generated module needn't be surjective: take $A=\mathbb Z$ and $T:\mathbb Z\to\mathbb Z:m\mapsto 2m$ !
Best Answer
I doubt that there is any better way than noting that the kernel is a direct summand of $M$, hence, finitely generated too. You could circumvent this a little by noting that if $f\colon A^n\to M$ is the map $e_i\mapsto x_i$ (the first component of your $\varphi$), then $\mathrm{id}_M-fg$ maps $M$ surjectively onto $\ker g$. Perhaps this feels a bit more concise.
In case you just don't want to deal with elements, here is a closely related argument in a slightly more abstract phrasing. Consider the short exact sequence $0\to \ker g\to M\to A^n\to 0 $ and the associated exact sequence after applying $\hom(-,\ker g)$: $$0\to\hom(A^n,\ker g)\to\hom(M,\ker g)\to\hom(\ker g,\ker g)\to \mathrm{ext}^1_A(A^n,\ker g).$$ Since $A^n$ is projective, $\mathrm{ext}^1_A(A^n,M)=0$ and so the map $\hom(M,\ker g)\to\hom(\ker g,\ker g)$, $\varphi\mapsto \varphi|_{\ker g}$, is surjective. Thus, there exists a morphism $\varphi\colon M\to \ker g$ such that $\varphi|_{\ker g}=\mathrm{id}_{\ker g}$. In particular, $\varphi$ is surjective and since $M$ is finitely generated, so is $\ker g$.