Let $F$ be a field, and $k=F[x,y]/(x^2,xy,y^2)$. Since $k$ is a finite-dimensional $F$-algebra, flat=projective, and for $k$-modules $M,N$, there is a natural isomorphism $\operatorname{Hom}_k(M,N^\ast)\cong (M\otimes_kN)^\ast$, where $L^\ast=\operatorname{Hom}_F(L,F)$ denotes $F$-dual.

So we have a counterexample if we can find a monomorphism $i:k\to Y$ and an epimorphism $p:Y'\to k^\ast$ of finite dimensional $k$-modules, and a homomorphism $\alpha:k\to k^\ast$ such that there is no homomorphism $\beta:Y\to Y'$ such that $\alpha=p\alpha i$ (i.e., the map $\operatorname{Hom}_k(Y,Y')\to\operatorname{Hom}_k(k,k^\ast)$ induced by $i$ and $p$ is not surjective).

Let $P=k$ and $I=k^\ast$ be the unique indecomposable projective and injective $k$-modules, and $F=k/(x,y)$ the unique simple $k$-module.

Let $i:k\to I(k)$ be the inclusion of $k$ into its injective hull (where $I(k)\cong k^\ast\oplus k^\ast$) and $p:P(k^\ast)\to k^\ast$ be the surjection from the projective cover of $k^\ast$ (where $P(k^\ast)\cong k\oplus k$.

Then there are homomorphisms $\alpha:k\to k^\ast$ such that the induced map $\operatorname{Hom}_k(F,k)\to\operatorname{Hom}_k(F,k^\ast)$ is non-zero, but there are no maps $\gamma:k^\ast\to k$ so that the induced map $\operatorname{Hom}_k(F,k^\ast)\to\operatorname{Hom}_k(F,k)$ is non-zero, so $\alpha$ gives a counterexample.

Translating back into the notation of the question, $f_1$ and $f_2$ are both the inclusion $i:k\to I(k)$.

The same works for any finite-dimensional commutative $F$-algebra $k$ that is not self-injective.

Let $F$ be a field, and $k=F[x,y]/(x^2,xy,y^2)$. Since $k$ is a finite-dimensional $F$-algebra, flat=projective, and for $k$-modules $M,N$, there is a natural isomorphism $\operatorname{Hom}_k(M,N^\ast)\cong (M\otimes_kN)^\ast$, where $L^\ast=\operatorname{Hom}_F(L,F)$ denotes $F$-dual.

So we have a counterexample if we can find a monomorphism $i:k\to Y$ and an epimorphism $p:Y'\to k^\ast$ of finite dimensional $k$-modules, and a homomorphism $\alpha:k\to k^\ast$ such that there is no homomorphism $\beta:Y\to Y'$ such that $\alpha=p\alpha i$ (i.e., the map $\operatorname{Hom}_k(Y,Y')\to\operatorname{Hom}_k(k,k^\ast)$ induced by $i$ and $p$ is not surjective).

Let $P=k$ and $I=k^\ast$ be the unique indecomposable projective and injective $k$-modules, and $F=k/(x,y)$ the unique simple $k$-module.

Let $i:k\to I(k)$ be the inclusion of $k$ into its injective hull (where $I(k)\cong k^\ast\oplus k^\ast$) and $p:P(k^\ast)\to k^\ast$ be the surjection from the projective cover of $k^\ast$ (where $P(k^\ast)\cong k\oplus k)$.

Then there are homomorphisms $\alpha:k\to k^\ast$ such that the induced map $\operatorname{Hom}_k(F,k)\to\operatorname{Hom}_k(F,k^\ast)$ is non-zero, but there are no maps $\gamma:k^\ast\to k$ so that the induced map $\operatorname{Hom}_k(F,k^\ast)\to\operatorname{Hom}_k(F,k)$ is non-zero, so $\alpha$ gives a counterexample.

Translating back into the notation of the question, $f_1$ and $f_2$ are both the inclusion $i:k\to I(k)$.

The same works for any finite-dimensional commutative $F$-algebra $k$ that is not self-injective.