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
In the spirit of first approach, there is a conjecture for a Clifford-algebra type proof of the 576-fold periodicity of TMF. This is a generalized cohomology theory constructed by piecing together all the elliptic cohomology theories together in a suitable way. I heard about this conjecture from Andre Henriques, who is working on a geometric approach to TMF using conformal nets.
The idea is that the free fermion conformal net (a introduction can be found in this article by Bartels, Douglas and Henriques) is to TMF as the Clifford algebras are to K-theory. For a suitable sense of equivalence, i.e. some generalization of Morita equivalence, $Free(n)$ and $Free(n+576)$ should be equivalent. I believe people are still far from a proof, but the motivation comes from looking at orientations: a manifold is orientable for K-theory if the frame bundle (a principal $SO(n)$-bundle) lifts to a principal $Spin(n)$-bundle. The $Spin(n)$ groups can be defined as a group in the Clifford algebra. For TMF, a manifold is orientable if the frame bundle extends to a principal $String(n)$-bundle, which can be obtained from $Spin(n)$ by killing $\pi_3$, just like $Spin(n)$ is obtained from $SO(n)$ by killing $\pi_1$. There is a way to define $String(n)$ using the free fermion conformal nets.
The only reference I know for these ideas is the following summary.