Given a topological space $X$ and a natural number $n$, to each rank $n$ real vector bundle $E\to X$ corresponds a principal $GL(n,\mathbb R)$-bundle $P(E)\to X$, called the frame bundle of $E$; the fiber $P(E)_x$ over a point $x\in X$ is the set of linear isomorphisms $p:\mathbb R^n\to E_x$. The group $GL(n,\mathbb R)$ acts on $P(E)$ on the right by $g: p\mapsto p\circ g$.
In the other direction, to a given a prinicipal $GL(n,\mathbb R)$-bundle $P\to X$ associate the vector bundle $E(P)=P\times\mathbb R^n/G$, where $G$ acts on $P\times \mathbb R^n$ diagonaly by $(p,v)\mapsto (pg,g^{-1}v)$.
You need to verify many small (and easy) details for $P(E)$ and $E(P)$ to be precisly and well defined. The upshot is that the maps $E\mapsto P(E)$ and $P\mapsto E(P)$ induce a natural bijection between the set of equivalence classes of rank $n$ real vector bundles over $X$ and the set of equivalence classes of principal $GL(n,\mathbb R)$-bundles over $X$.
You can modify the above easily to complex vector bundles and manifolds.
So we see that vector bundles (real or complex) correspond to principal $G$-bundles, where $G$ is the general linear group (real or complex). You can extend the correpondence to include other groups $G$, by adding more structure on the vector bundle (orientation, euclidean metric etc).
This may give the impression that working with vector bundles or principal bundles is "basicaly the same", perhaps a matter of taste. This is not the case, and both concepts are essential. But this will take too long to explain, perhaps the subject of a seperate question (sorry).
The Borel subgroup $B(n)$ of $Gl(n)$ is isomorphic to the group of upper triangular matrices which preserves a family of vector supspaces $V_i\subset V_{i+1}$ and dimension of $V_i=i$, this implies that if $p:P\rightarrow M$ has a $B(n)$ reduction defined by $(U_j,g_{jk}$, $g_{jk}$ preserves $V_i$ and defines a flag of vector bundles $p_i:P_i\rightarrow M$ whose typical fibre is $V_i$.
Best Answer
Short answer: $GL$ does not exhaust all possible Lie groups. A vector bundle always has an associated principal bundle with fiber $GL$ (or a subgroup). In particular, some Lie groups do not admit a faithful linear representation (see this question on MO). Therefore, they cannot be associated to any vector bundle (in a way, no vector bundle "represents them well").
The other way around, given a vector bundle, one can always find an associated principal bundle, as you probably know. But for many theoretical problems, the vector bundle is the most natural object to work with (sometimes, simply because it is a nice module/vector space). For example, on a vector bundle both the action of the group and the sum of vectors are natural operations. Instead, on a principal bundle, the sum is not natural. For principal bundles associated with vector bundles you can try to inherit the sum from the vector space, but it is not guaranteed that summing two sections you get again a section - can you see why? (think of a $U(1)$-bundle...)