I actually don't think that this view of light being in a quantum superposition is anything new: what Discover magazine is describing (I believe) is the stock standard picture of how one would describe a system of cells, molecules, chloroplasts, fluorophores, whatever interacting with the quantised electomagnetic field.
My simplified account here (answer to Physics SE question "How does the Ocean polarize light?") addresses a very similar question. The quantised electromagnetic field is always in superposition before the absorption happens and, as light reaches a plant, it becomes a superposition of free photons and excited matter states of many chloroplasts at once.
To learn more about this kind of thing, I would recommend
M. Scully and M. Zubairy, "Quantum Optics"
Read the first chapter and the mathematical technology for what you are trying to describe is to be found in chapters 4, 5 and 6.
The truth is, photons do not bounce from cell to cell like ping pong balls. So that theory happens to be incorrect.
Further questions and Edits:
But this is about the energy FROM the photon... Would whatever you are saying still work for that? Plus, I would like to see some math...
Energy is simply a property of photons (or whatever is carrying it): there has to be a carrier to make any interaction happen. All interactions we see are ultimately described by this. See eq (1) and (2) here, this is for the reverse process (emission) but you are ultimately going to write equations like this. To get a handle on this quickly look into this Wikipedia article (Quantization of the electromagnetic field) and then read Chapter 1 from Scully and Zubairy.
Ultimately, you're going to need to write down a one-photon Fock state, and add to the superposition excited atom states. The neater way to do this is with creation operators acting on the universal, unique quantum ground state $\left|\left.0\right>\right.$: we define $a_L^\dagger(\vec{k},\,\omega),\,a_R^\dagger(\vec{k}\,\omega)$ to be the creation operators for the quantum harmonic oscillators corresponding to left and right handed plane waves with wavenumber $\vec{k}$ and frequency $\omega$. Then a one-photon state in the oscillator corresponding to the classical solution of Maxwell's equation with complex amplitudes $A_L(\vec{k},\,\omega), A_R(\vec{k},\,\omega)$ in the left and right handed classical modes is:
$$\left|\left.\psi\right>\right.=\int d^3k\,d\omega\left(A_L(\vec{k},\,\omega)\,a_L^\dagger(\vec{k},\,\omega)+A_R(\vec{k},\,\omega)\,a_R^\dagger(\vec{k}\,\omega)\right)\,\left|\left.0\right>\right.$$
To define an absorption, Scully and Zubairy show that the probability amplitude for an absorption at time $t$ and position $\vec{r}$ is proportional to:
$$\left<\left.0\right.\right| \hat{E}^+(\vec{r},t)\left|\left.\psi\right>\right.$$
where $\hat{E}$ is the electric field observable and $\hat{E}^+$ its positive frequency part (the part with only annihilation operators and all the creation operators thrown away).
Alternatively you can in principle model absorption by writing down the Hamiltonian which is going to look something like:
$$\int d^3k\,d \omega\left(a_L^\dagger(\vec{k},\,\omega)\,a_L^\dagger(\vec{k},\,\omega)+a_R^\dagger(\vec{k}\,\omega)\,a_R(\vec{k},\,\omega) \right)+\sum\limits_{\text{all chloroplasts }j} \int d^3k\,d\omega\,\sigma^\dagger_j\left(\kappa_{j,L}(\vec{k},\,\omega)\,a_L(\vec{k},\,\omega)+\kappa_{j,L}(\vec{k},\,\omega\,a_R(\vec{k},\,\omega) \right)+\\\sum\limits_{\text{all chloroplasts }j} \int d^3k\,d\omega\,\left(\kappa_{j,L}(\vec{k},\,\omega)\,a^\dagger_L(\vec{k},\,\omega)+\kappa_{j,L}(\vec{k},\,\omega)\,a^\dagger_R(\vec{k},\,\omega) \right)\sigma_j$$
where $\sigma_j^\dagger$ is the creation operator for a raised chlorophore at site $j$ and the $\kappa$s measure the strength of coupling.
This is complicated stuff and takes more than a simple tutorial to write down.