Here is my attempt to address Eric's actual question. Given a real $n$-dimensional vector bundle $E$ on a space $X$, there is an associated Thom space that can be understood as a twisted $n$-fold suspension $\Sigma^E X$. (If $E$ is trivial then it is a usual $n$-fold suspension $\Sigma^n X$.) In particular, if $E=L$ is a complex line bundle, it is a twisted double suspension. In particular, if $X = \mathbb{C}P^\infty$, the twisted double suspension of the tautological line bundle $L$ satisfies the equation
$$\Sigma^L\mathbb{C}P^\infty = \mathbb{C}P^\infty.$$
As I understand it, Eric wants to know whether this periodicity can be interpreted as a Bott map, maybe after some modification, and then used to prove Bott periodicity. What I am saying matches Eric's steps 1 and 2. Step 3 is a modification to make the map look more like Bott periodicity.
I think that the answer is a qualified no. On the face of it, Eric's map does not carry the same information as the Bott map. Bott periodicity is a theorem about unitary groups and their classifying spaces. What Eric has in mind, as I understand now, is a result of Snaith that constructs a spectrum equivalent to the Bott spectrum for complex K-theory by modifying $\mathbb{C}P^\infty$. Snaith's model has been called "Snaith periodicity", but the existing arguments that it is the same are a use and not a proof of Bott periodicity. (In that sense, Snaith's model is stone soup, although that metaphor is not really fair to his good paper.)
For context, here is a quick definition of Bott's beautiful map as Bott constructed it in the Annals is beautiful. In my opinion, it doesn't particularly need simplification. The map generalizes the suspension relation $\Sigma S^n = S^{n+1}$. You do not need Morse theory to define it; Morse theory is used only to prove homotopy equivalence. Bott's definition: Suppose that $M$ is a compact symmetric space with two points $p$ and $q$ that are connected by many shortest geodesics in the same homotopy class. Then the set of these geodesics is another symmetric space $M'$, and there is an obvious map $\Sigma M' \to M$ that takes the suspension points to $p$ and $q$ and interpolates linearly. For example, if $p$ and $q$ are antipodal points of a round sphere $M = S^{n+1}$, the map is $\Sigma(S^n) \to S^{n+1}$. For complex K-theory, Bott uses $M = U(2n)$, $p = q = I_{2n}$, and geodesics equivalent to the geodesic $\gamma(t) = I_n \oplus \exp(i t) I_n$, with $0 \le t \le 2\pi$. The map is then
$$\Sigma (U(2n)/U(n)^2) \to U(2n).$$
The argument of the left side approximates the classifying space $BU(n)$. Bott show that this map is a homotopy equivalence up to degree $2n$. Of course, you get the nicest result if you take $n \to \infty$. Also, to complete Bott periodicity, you need a clutching function map $\Sigma(U(n)) \to BU(n)$, which exists for any compact group. (If you apply the general setup to $M = G$ for a simply connected, compact Lie group, Bott's structure theorem shows that $\pi_2(G)$ is trivial; c.f. this related MO question.)
At first glance, Eric's twisted suspension is very different. It exists for $\mathbb{C}P^\infty = BU(1)$, and of course $\mathbb{C}P^\infty$ is a $K(\mathbb{Z},2)$ space with a totally different homotopy structure from $BU(\infty)$. Moreover, twisted suspensions aren't adjoint to ordinary delooping. Instead, the space of maps $\Sigma^L X \to Y$ is adjoint to sections of a bundle over $X$ with fiber $\mathcal{L}^2 Y$. The homotopy structure of the twisted suspension depends on the choice of $L$. For instance, if $X = S^2$ and $L$ is trivial, then $\Sigma^L S^2 = S^4$ is the usual suspension. But if $L$ has Chern number 1, then $\Sigma^L S^2 = \mathbb{C}P^2$, as Eric computed.
However, in Snaith's paper all of that gets washed away by taking infinitely many suspensions to form $\Sigma_+^{\infty}\mathbb{C}P^\infty$, and then as Eric says adjoining an inverse to a Bott element $\beta$. (I think that the "+" subscript just denotes adding a disjoint base point.) You can see what is coming just from the rational homotopy groups of $\Sigma^\infty \mathbb{C}P^\infty$. Serre proved that the stable homotopy of a CW complex $K$ are just the rational homology $H_*(K,\mathbb{Q})$. (This is related to the theorem that stable homotopy groups of spheres are finite.) Moreover, in stable, rational homotopy, twisted and untwisted suspension become the same. So Snaith's model is built from the fact that the homology of $\mathbb{C}P^\infty$ equals the homotopy of $BU(\infty)$. Moreover, there is an important determinant map
$$\det:BU(\infty) \to BU(1) = \mathbb{C}P^\infty$$
that takes the direct sum operation for bundles to tensor multiplication of line bundles. Snaith makes a moral inverse to this map (and not just in rational homology).
Still, searching for a purely homotopy-theoretic proof of Bott periodicity is like searching for a purely algebraic proof of the fundamental theorem of algebra. The fundamental theorem of algebra is not a purely algebraic statement! It is an analytic theorem with an algebraic conclusion, since the complex numbers are defined analytically. The best you can do is a mostly algebraic proof, using some minimal analytic information such as that $\mathbb{R}$ is real-closed using the intermediate value theorem. Likewise, Bott periodicity is not a purely homotopy-theoretic theorem; it is a Lie-theoretic theorem with a homotopy-theoretic conclusion. Likewise, the best you can do is a mostly homotopy-theoretic proof that carefully uses as little Lie theory as possible. The proof by Bruno Harris fits this description. Maybe you could also prove it by reversing Snaith's theorem, but you would still need to explain what facts you use about the unitary groups.
(The answer is significantly revised now that I know more about Snaith's result.)
As Thorny said, Milnor's axiomatic definition seems to be precisely the best way of proving that different definitions are the same. The main thrust of his "definition" is the proof that any invariants that satisfy these axioms must be the same as Stiefel-Whitney classes. In his book, they connect the two notions I describe below as well as the Steenrod-squares definition. They should also serve to prove that all the definitions you talk about are the same.
The rest of this answer might have less to do with your exact question than with my tendency to see an interesting question title and start writing. Sorry! Still, I feel that they are things that should be said (or, at least, don't deserve to be deleted).
I think there are two very important ways to understand characteristic classes. Both are explained in Milnor's Characteristic Classes, but not as the definition, since they are not as precise (but, to me, they are much more intuitive).
Think of your vector bundle as a map from your space X into a Grassmanian. The cohomology of the Grassmanian (more precisely, either the $\mathbb Z/2$ cohomology of the real Grassmanian, or the usual cohomology of the complex Grassmanian) is a polynomial algebra on some generators. The characteristic classes (Stiefel-Whitney or Chern, respectively) are precisely the pullbacks of these cohomology classes to X via the map.
Reading your question carefully, I guess you already knew this. Still, I think you should give this definition more credit. In particular, I think that this is the best explanation of the philosophical reason why "characteristic classes" exist. On thing that confuses me: why are the pullbacks of the integer cohomology of the real Grassmanian never called characteristic classes? I'm sure they are a pain to calculate, but that doesn't justify why nobody seems to care for them at all...
You can understand them through obstruction theory (another reference: Steenrod's "Theory of Fibre Bundles). The idea is to generalize the definition of the Euler characteristic using vector fields. Namely, try to construct a nowhere-zero section of your bundle. The obstruction will be a cohomology class, which is called the Euler class (and corresponds mod 2 to the top Stiefel-Whitney class). Try to construct two linearly independent nowhere-zero sections of the bundle. The obstruction will be a cohomology class which, mod 2, will be the next (one dimension lower) Stiefel-Whitney class. If you keep going like this, you'll construct all the classes.
Here is an explanation of why the obstructions to constructing non-zero sections are cohomology classes, for the case of a single section.
Think of your space X as a CW-complex; start constructing it on the 0-skeleton, and then try extending the section to 1-skeleton, and so on. At each step, you will basically be solving the following problem:
Given a vector field on the boundary $S^{n-1}$ of the ball $B^n$, can you extend it to the whole ball?
To solve this, think of the vector field as a map $S^{n-1}\to \mathbb R^m$ where m is the dimension of your bundle (you can assume that the bundle is trivial over the ball $B^n$ since the ball is contractible). Since the vector field is supposed to be nowhere zero, you can think of this as a map $S^{n-1}\to S^{m-1}$. If $n<m$, this map is always nullhomotopic and always extends to the ball. If $n=m$, you get an integer, the degree of the map, which tells you if you can extend. Since you get an integer for each degree-m cell of the CW-complex, you get something that looks like a cohomology class in $H^m(X)$ (of course, you need to verify separately that it actually is one, and if you are precise enough, you'll see that these integers only make sense mod 2). This is the Euler class.
If you wanted to construct two linearly independent sections, first construct one up to the $n-1$-skeleton (which is always possible). Now, let's start making the second one. You might as well require the second section to be orthogonal to the first. So, in the extension problem, you'll have a map $S^{n-1}\to \mathbb R^{m-1}$ where the $\mathbb R^{m-1} \subset \mathbb R^m$ is the subspace orthogonal to the first section. Since it also can't be zero, it's really a map $S^{n-1}\to S^{m-2}$. The rest of the argument is the same; you get a class in $H^{m-1}(X)$.
Usual disclaimer: there may be mistakes anywhere. Please point them out!
Best Answer
That ain't easy. However, if you accept that DeRham cohomology is nothing but the cohomology of a particular soft resolution of the constant sheaf, then the book Cohomology of Sheaves by B. Iversen is of great help, albeit a bit hard to digest. The techniques in this book are really powerful and they've gotten me out of a jam many a times.