Let $M$ be an $R$-module. The flat cover of $M$ (which always exists) can be defined as a pair $(F, \phi)$ such that $\phi: F \to M$ is surjective, $F$ is a flat module, and every homomorphism from a flat module to $M$ factors through $F$. Moreover, any map from $F$ to itself that commutes with the map to $M$ is an automorphism of $F.
Okay, this is just the definition of a flat cover. The question is whether one can define a projective cover by simply replacing the last definition, replacing $F$ with $P$, where $P$ is a module in the class of all $R$-projective modules. It can be proven that this is equivalent in the case of a projective cover. In other words, "$(P, \phi)$" is a projective cover of an $R$-module if $\phi: P \to M$ is a superfluous epimorphism, meaning that for every $L \leq P$ such that $Ker(\phi) + L = P$, then $L = M$; in other words, $Ker(\phi)$ is superfluous in $P$.
A proof showing the equivalence of these two ways to define a projective cover can be found in "Flat covers" by Xu on page 15, as Theorem 1.2.12.
So, the question is, can we define a flat cover in a similar way? That is, can we say the following: $(F, \phi)$ is a flat cover of an $R$-module $M$ if $\phi: F \to M$ is a superfluous epimorphism and $F$ is a flat module? My intuition says no, but I find it strange that nobody uses this definition, although I cannot find a counterexample. I also have a hunch that this is not always true. Ideally, I would like to come across an example where $Ker(\phi)$ is not superfluous in $F$ while $F$ is flat. Thanks in advance for reading and for your help.
The alternative definition is not equivalent for noncommutative rings. (Although the question doesn't specify commutativity, it has the "commutative-algebra" tag, so possibly that's what the OP intended.)
The "superfluous" definition of covers is studied in
Jain, S. K.; López-Permouth, Sergio R.; Rowen, Louis H., Superfluous covers, Commun. Algebra 23, No. 5, 1663-1677 (1995). ZBL0829.16001.
(PDF available here.)
It's shown there that not every module has a flat cover in the "superfluous" sense, and so, since every module does have a flat cover in the usual sense, the two definitions must be nonequivalent.
However, it looks as though the argument in the paper necessarily needs a ring that is noncommutative. I don't know what happens for commutative rings, but I would guess the two definitions are still nonequivalent.