Let $G$ be a topological group (for sake of simplicity I think it suffice to assume that $G$ is finite group. It is well known that for any homotopy class of a map $\phi: X \to BG$ (with $X$ nice 'enough'; say paracompact or already a CW complex) the pullback of universal prinicipal $G$-bundle $u: EG \to BG$ by $\phi^*$ induces a natural bijection between isomorphism classes of principal $G$-bundles over $X$ $\mathcal{P}_G(X)$ and homotopy classes $[X, BG]$.
Now I would like to understand what is the intuitive reason which the universal bundle $u: EG \to BG$ has this 'miracle' property. The key seems to sitting in the explicit comstruction of $EG$ and $BG$. The most 'constructive' I know is due to Milnor and here is a discussion where this construction is explicitely described
The key point I would like to use now is that it is known that every $G$-bundle over a nice enough base space is completely described by an associated $1$-cocycle (see below) and I thing that the reason why $u: EG \to BG$ has this miracle property is somehow reflected in the structure of it's associated $1$-cocycle. Let recall the most important properties of this formalism:
Let $p:E \to X $ a $G$-fiber bundle with fiber $F$ (say a finite set with transitive and free $G$-action; in our case later it will be $G$! itself). Let $\mathcal{U}= \{U_i \}_i$ covering of $X$ over which $E$ trivializes, that is we have
(1) a colection of homeomorphisms $\psi_i: U_i \times F \to p^{-1}(U_i)$ compatible with projection $p$
(2) the $1$-cycle (ie the set of transition functions) $g_{ij}: U_j \cap U_i \to G$ such that for every $(x, f) \in (U_j \cap U_i) \times F$ we have
$$ \psi_j^{-1} \circ \psi_i (x,f) = (x, (g_{ji}(x))(f))$$
(3) the transition functions satisfy $g_{ii} = e_G$ (constant function), $g_{ji}= g_{ij}^{-1}$ and $g_{kj} \cdot g_{ji}= g_{ki}$
Moreover two $G$-bundles $E_1, E_2$ over same base $X$ are isomorphic if there exist fine enough covering $\{U_i \}_i$ of $X$ such that there eist a family $g_i: U_i \to G$ with
$$ g^2_{ij} = g_i \cdot g^1_{ij} \cdot g_j^{-1} $$
where $\{g^1_{ij}\}$ and $\{g^2_{ij}\}$ are $1$-cycles of $E_1$ and $E_2$.
So our task is to check that for every principal $G$-bundle $E \to X$ with cover $\{U_i\}_i$ and $1$-cycle $\{g^E_{ij}\}$ there exist a and contunuous map $f:X \to BG$ and a covering $\{V_i\}_i$ of $BG$ with $1$-cycle $\{g^u_{ij}\}$ fine enough such that it induces via taking preimages a finer cover $\{ f^{-1}(V_i)\}_i$ than $\{ U_i \}_i$, such that the refinened $1$-cycle $\{\overline{g}^E_{ij}\}$ over $\{f^{-1}(V_i)\}_i$ differs from pullback $1$-cycle $\{(g^u \circ f)_{ij}\}$ only by a conjugation by the family $g_i: f^{-1}(V_i) \to G$ such that
$$ \overline{g}^E_{ij} = g_i \cdot (g^u \circ f)_{ij} \cdot g_j^{-1} $$
How to prove it? Or let me ask a relataded quetion. In which sense are the $1$-cycles $EG$ of more 'flexiblel' such we can hit every $1$-cocycle via pullback?