Here is one possible way of answering the question, using the simplicial model for $EG$. In this context, we can show very explicitly that the map $\Sigma X\to BG$ built (via adjointness) from the clutching function for $E$ classifies $E$ (and then if $X$ is a CW complex, any other classifying map for $E$ is homotopic to this one, by a standard cell-by-cell argument).

If $E\to \Sigma X$ is a principal $G$-bundle, then as explained in the question, $E$ is trivial over each half of $\Sigma X$. So $E$ is (isomorphic to) the quotient space
$$(C_{-} X\times G) \cup (C_+ X \times G)\,/\,(x_-, g)\sim(x_+, c(x)g)$$
where $c: X\to G$ is the clutching function, and the notation for the relation is supposed to indicate that $x_-$ and $x_+$ are the copies of $x$ in the bases of the cones $C_-X$ and $C_+ X$ (respectively).

Let's assume that $G$ is a Lie group, so that $BG$ can be taken to mean the bar construction on $G$ (as in, say, "Segal's paper Classifying spaces and spectral sequences"). Then there is a natural map $\Sigma G \to BG$ given by the fact that $G$ is the space of 1-simplices in $BG$, and it has an adjoint $i: G\to \Omega BG$ (which sends $G$ to the loop that "runs around the 1-simplex labeled '$g$'," so to speak).

The function $i \circ c: X \to \Omega BG$ has an adjoint $f: \Sigma X \to BG$. It sends $x$ to $[c(x), t]$, which means "the point t units along the 1-simplex labeled $c(x)$." We want to prove that $f$ classifies $E$ - that is, we want to construct a map $\tilde{f}: E\to EG$ that is $G$-equivariant and covers $f$.

This can be done very explicitly. In Segal's setup, $EG$ is also a simplicial space, with 1-simplices parametrized by $G\times G$; the $G$-action on this (universal) principal bundle is induced by right-multiplication in $G$, and the map $EG\to BG$ sends the 1-simplex $(g,h)$ to $g h^{-1}$.

I want to think of $\Sigma X$ as $X\times I$ with both ends collapsed to points, so I'll take $C_-$ to be parametrized by $t\in [0,1/2]$ and $C_+$ to be parametrized by $t\in [1/2, 1]$.

We can define $\tilde{f}$ separately on the two halves of $E$: on the left half
$$C_{-} X\times G = \left[(X\times [0,1/2])/(X\times 0)\right]\times G,$$
we set $\tilde{f}(x, t, g) = [(c(x)g, g), t]$. On the right half
$$C_{+} X\times G = \left[(X\times [1/2, 1])/(X\times 1)\right]\times G,$$
we set $\tilde{f}(x, t, g) = [(g, c(x)^{-1}g), t]$. Note that in both cases, when we project down to $BG$ we get just $[c(x), t]$, and that the map is well-defined on $E$ because when $t=1/2$ (where the two cones meet) the first formula sends $(x_-, g) = (x, 1/2, g)$ to $[(c(x)g, g), 1/2]$ and the second formula sends the equivalent point $(x_+, c(x)g) = (x, 1/2, c(x)g)$ to $[(c(x)g, c(x)^{-1} c(x)g], 1/2] = [(c(x)g, g), 1/2]$.

I guess something similar probably works for any topological group $G$, if one uses Milnor's join construction instead of Segal's simplicial model, but I haven't thought through it.