Let $A \to B$ be a homomorphism of commutative rings. I would like to find a criterion for the flatness of $A \to B$ which does not involve the notion of kernels; it should rather involve cokernels. This is partially motiviated by this question. I suggest the following: Consider for $A$-modules $M,N$ the canonical homomorphism of $B$-modules
$\alpha_{M,N} : \mathrm{Hom}_A(M,N) \otimes_A B \to \mathrm{Hom}_B(M \otimes_A B, N \otimes_A B) \cong \mathrm{Hom}_A(M,N \otimes_A B).$
For fixed $N$, the class $\{M : \alpha_{M,N} \text{ is an isomorphism}\}$ contains $A$ and is closed under finite direct sums. Now consider the following statement:
($\dagger$) For all $A$-modules $N$, the set $\{M : \alpha_{M,N} \text{ is an isomorphism}\}$ is closed under finite colimits, or equivalently, under cokernels.
Note that $\dagger$ implies, in particular, that $\alpha_{M,N}$ is an isomorphism whenever $M$ is finitely presented. But probably $\dagger$ is a stronger statement.
If $A \to B$ is flat, then $\dagger$ holds since then $\alpha_{-,N} : \mathrm{Hom}_A(-,N) \otimes_A B \to \mathrm{Hom}_A(-,N \otimes_A B)$ is a natural transformation between right exact functors $\mathrm{Mod}(A) \to \mathrm{Mod}(B)^{\mathrm{op}}$. What about the converse?
Assume that $\dagger$ holds. Then for all $N$ and all exact sequences $M'' \to M \to M' \to 0$ of modules over $A$ we have the following commutative diagram:
$$\begin{matrix} 0 & & 0 \\\ \downarrow & & \downarrow \\\ \mathrm{Hom}_A(M',N) \otimes_A B & \rightarrow & \mathrm{Hom}_A(M',N \otimes_A B) \\\ \downarrow & & \downarrow \\\ \mathrm{Hom}_A(M,N) \otimes_A B & \rightarrow & \mathrm{Hom}_A(M,N \otimes_A B) \\\ \downarrow & & \downarrow \\\ \mathrm{Hom}_A(M'',N) \otimes_A B & \rightarrow & \mathrm{Hom}_A(M'',N \otimes_A B) \end{matrix}$$
The right column is exact anyway. $\dagger$ says that we have an isomorphism in the first row as soon we have it in the two lower rows. This shows in particular that $- \otimes_A B$ preserves a large class of monics, namely those of the form $\mathrm{Hom}(M',N) \to \mathrm{Hom}(M,N)$ where $M \twoheadrightarrow M'$ and $M,M'$ belong to the class (for example, if they are finitely presented). Perhaps this is a first step in order to show that $A \to B$ is flat.