Let $A$ be an associative, unital, commutative, $\mathbb{K}$-Algebra. Let $V$ and $U$ be $A$-modules and $f:V\rightarrow U$ a $\mathbb{K}$-linear map. Consider $A$ as an $A$-module over itself. Suppose for every $A$-module homomorphism $\phi:U\rightarrow A$ the composition $\phi\circ f:V\rightarrow A$ is an $A$-module homomorphism.
Is it true that $f$ must be in fact an $A$-module homomorphism?
(Of most interest is the case $\mathbb{K}=\mathbb{R}$)
Take $U=V= \mathbb{C}[x]/\langle x \rangle$ and $A=\mathbb{R}[x]$. Clearly, $U, \ V$ are $A$-modules and $A$ is an algebra over $\mathbb{R} $ having all the properies you asked for. Following the comment of Mohan we show that there no none-zero $A$-module homomorphism from $U$ to $A$. Let $\varphi: U\rightarrow A$ be an $A$-module homomorphism. Then we have for $[a]\in U$ $$ x \cdot \varphi([a])=\varphi([ax])=\varphi([0])=0.$$ As $A$ is integral we get $\varphi([a])=0$. Thus, $\varphi$ is the zero map.
Define the map $f: U \rightarrow V$ in the following way, for $p,q\in \mathbb{R} [x]$ we define $f([p +iq])=[p].$ This is a well-defined $\mathbb{R}$-module homomorphism, but it is not an $A$-module homomorphism as $$f(i[i])=f([-1]) = [-1] \neq [0] =i [0] = i f([i])$$