The following is thoroughly useless general nonsense. Its main problem is that it lacks geometry. Even so, it kind of does the job and it is probably worth mentioning.
Let $G$ be the structure group of your vector bundle. For simplicity, I will assume that $G$ is connected (for example, $G$ could be $U(n)$ or $SO(n)$). Let our vector bundle be classified by a map $f: X\to BG$.
Definition: Let $k \ge 1$ be an integer. A $k$-structure is a pair $(Z,g)$ such that
$\bullet$ $Z$ is a path connected space.
$\bullet$ $Z$ has vanishing homotopy groups above dimension $k$, and
$\bullet$ $g: Z\to BG\, $ is a map.
[To such pairs $(Z_i,g_i)$ for $i=0,1$ and are equivalent there is a (finite chain of) weak homotopy equivalences over $BG$ from $Z_0$ to $Z_1$.]
Example: A 1-structure amounts to a discrete group $H$ and a map $BH\to BG$.
Definition: A $k$-flat reduction of $f$ consists of a $k$-structure $(Z,g)$ and a factorization of $f$ up to homotopy as $\require{AMScd}$ $$ \begin{CD} X \to Z @>g>> BG\, . \\ \end{CD} $$
Examples:
(1). A 1-flat reduction of $f$ is the same thing as a flat reduction of the associated vector bundle.
(2). An oriented line bundle over $X$ automatically admits a $2$-flat reduction (this is a tautology, since $BSO(2) = K(\Bbb Z,2)$. In particular, the canonical line bundle over $\Bbb CP^1$ admits a 2-flat reduction but not a 1-flat reduction (since it isn't flat). More generally, a finite Whitney sum of oriented line bundles admits a $2$-flat reduction, since $BSO(2)^{\times j} = K(\Bbb Z^j,2)$.
(3). I don't know of examples in degrees $\ge 2$. However, if one is willing to work instead in the rational homotopy category, there are examples in higher degrees. Here's one: let $X= S^4$. This includes into $BS^3 = BSU(2)$. Rationally, this is a $K(\Bbb Q,4)$. Taking $G = SU(2)$ and $f: X\to BG$ to be the inclusion (which represents the quaternionic line bundle), we see that $f$ is rationally $4$-flat but not rationally $3$-flat.