This essentially follows from the results of
Auslander, M.; Smalø, Sverre O., Preprojective modules over Artin algebras, J. Algebra 66, 61-122 (1980). ZBL0477.16013.
and
Auslander, M.; Smalø, Sverre O., Almost split sequences in subcategories, J. Algebra 69, 426-454 (1981). ZBL0457.16017.
but an explicit statement and proof can be found in
Smalø, Sverre O., Torsion theories and tilting modules, Bull. Lond. Math. Soc. 16, 518-522 (1984). ZBL0519.16016.
Concretely, if $\mathcal{T}$ is a functorially finite torsion class and
$$A\stackrel{\alpha}{\longrightarrow} T_0\longrightarrow T_1\longrightarrow0$$
is an exact sequence with $\alpha$ a minimal left $\mathcal{T}$-approximation, then the indecomposable Ext-projective modules are the indecomposable summands of $T_0\oplus T_1$.
There are several parts to computing this, since one should first understand both constructions $\mathrm{Hom}_A(M,A)$ and $DN$, which are best thought of in terms of $A$-modules, and then we want to turn $D\mathrm{Hom}_A(M,A)$ back into a quiver representation. I will go through these constructions in general, before answering your specific problem at the end.
If $M$ is a left $A$-module, then the vector space $\mathrm{Hom}_A(M,A)$ becomes a right $A$-module via
$$ fa \colon M\to A, \quad m\mapsto f(m)a. $$
Let $e_i$ be the idempotent corresponding to vertex $i$. Then $Ae_i=P_i$ is the indecomposable projective $A$-module corresponding to vertex $i$, and
$$ \mathrm{Hom}_A(M,A)e_i = \mathrm{Hom}_A(M,Ae_i) = \mathrm{Hom}_A(M,P_i). $$
If $\alpha\colon i\to j$ is an arrow, then this yields an $A$-module map
$$ \alpha^\ast \colon P_j \to P_i, \quad a \mapsto a\alpha \quad\textrm{for $a$ in }P_j=Ae_j. $$
We therefore get an induced linear map
$$ \alpha^\ast \colon \mathrm{Hom}_A(M,P_j) \to \mathrm{Hom}_A(M,P_i), \quad f\mapsto\alpha^\ast f, \textrm{ which sends $m$ to }f(m)\alpha. $$
We next consider the vector space duality $D$. If $N$ is a right $A$-module, then $DN$ is a left $A$-module via
$$ a\xi\colon N\to k, \quad n\mapsto\xi(na). $$
In particular, $e_iDN=D(Ne_i)$. For an arrow $\alpha\colon i\to j$ we have the linear map $Ne_j\to Ne_i$, $n\mapsto n\alpha$, and then the induced linear map
$$ \alpha \colon D(Ne_i) \to D(Ne_j), \quad \xi\mapsto\alpha\xi, \textrm{ which sends $n\in Ne_j$ to $\xi(n\alpha)$.} $$
(Note that this is where the arrows get reversed: $\alpha\colon i\to j$ yields a map $Ne_j\to Ne_i$ for a right $A$-module $N$.)
We now put this together. For a left $A$-module $M$, the quiver representation corresponding to $D\mathrm{Hom}_A(M,A)$ has the vector space $D\mathrm{Hom}_A(M,P_i)$ at vertex $i$, and for an arrow $\alpha\colon i\to j$, we have the linear map
$$ \alpha\colon D\mathrm{Hom}_A(M,P_i) \to D\mathrm{Hom}_A(M,P_j), \quad \xi\mapsto\alpha\xi, \textrm{ which sends $g\in\mathrm{Hom}_A(M,P_j)$ to $\xi(\alpha^\ast g)$}. $$
OK. So what does this do in your special case, where $M=P_2$? We have $\mathrm{Hom}_A(M,P_2)=k\,\mathrm{id}$, and $\mathrm{Hom}_A(M,P_1)=k\,\alpha^\ast$, where $\mathrm{id}$ is the identity map and $\alpha\colon 1\to 2$ is the arrow.
Let $\xi$ and $\eta$ be the respective dual bases, so that $D\mathrm{Hom}_A(M,A)$ is one dimensional at each vertex, with basis $\eta$ at vertex 1 and $\xi$ at vertex 2.
Finally, $\alpha\eta(\mathrm{id})=\eta(\alpha^\ast)=1=\xi(\mathrm{id})$, so $\alpha$ sends $\eta$ to $\xi$, and we have the quiver representation $k \xrightarrow{1} k$ as expected.
Best Answer
I think I just missed the obvious.
The key point is that the projective dimension is bounded and thus choosing m larger than the Finitistic dimension of $A^{op}$ will yield that at some point (at most after m steps) all the morphisms in the resolution will split, so in particular $f_0 $ will be a split monomorphism.