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.)
Best Answer
Homology groups and homotopy groups are two sides of the same story. Homotopy groups tell us all the ways we can have a map Sn → X, and in particular describe all ways we can attach a new cell to our space. On the other side, the homology groups of a space change in a very understandable way each time we attach a new cell, and so they tell us all the ways that we could build a homotopy-equivalent CW-complex. In cases where we can understand both of them, we can get things like complete theorems about classification of spaces.
Here's an example where we can compute: classification of the homotopy types of compact, orientable, simply-connected 4-manifolds. (I originally saw this is Neil Strickland's bestiary of topological spaces.)
Poincare duality tells us that the homology groups are finitely generated free in degree 2, ℤ in degrees 0 and 4, and zero elsewhere. We can cut out a closed ball, and get an expression of the manifold as obtained from a manifold-with-boundary N by attaching a 4-cell. The Hurewicz theorem tells us that we can construct a map from a wedge of copies of S2 to N which induces an isomorphism on H
*, and by the (homology) Whitehead theorem this is a homotopy equivalence. So our original manifold is obtained, up to homotopy equivalence, by attaching a 4-cell to $\bigvee S^2$.How many ways are there to do this? It is governed by $\pi_3 (\bigvee S^2)$, which we can compute because it's low down enough. This homotopy group is naturally identified with the set of symmetric bilinear pairings $H^2(\bigvee S^2) \to \mathbb{Z}$, and this identification is given by seeing how the cup product acts after you attach a cell. So these 4-manifolds are classified up to homotopy equivalence by the nondegenerate symmetric bilinear pairing in their middle-dimensional cohomology.
Some of what we used here is general and well-understood machinery about homology, homotopy, and their relationships. Wouldn't it be nice if the standard tools were always so effective? But the real meat is that we have a complete understanding of homology and homotopy in the relevant ranges. It turns our questions about classification into questions about pure algebra. For questions that require specific knowledge about higher homotopy groups of spheres (or even lower homotopy groups of complicated spaces), it is much harder to get answers. There aren't a lot of spaces where we have complete understanding of both the homology groups and the homotopy. We have tools for reconstructing the former from the latter but their effectiveness wears down the farther out you try to go.
There are categories that are somewhat like the homotopy category of spaces where we can get an immediate and specific understanding of both sides of the coin.
One such example is the category of chain complexes over a ring R. There, our fundamental building block is R itself. The homology of any chain complex tells us both how R can be mapped in modulo chain homotopy, and how complicated any construction of the underlying chain complex must be. A more complicated example would be the category of differential graded modules over a DGA, where the divide between how things can be constructed and how things can have new cells attached is, at the very least, governed by the complexity of H
*A as a ring, and then by the secondary algebraic operations if A is far from being anything like formal.Another such example is the rational homotopy theory of simply-connected spaces you mentioned. There, homology and homotopy are roughly something like the difference between a ring's underlying abelian group structure and how you build it using generators and relations.
So you might think of the complexity of homotopy groups as telling us how much more complicated spaces are than chain complexes.