In calculating the electron dispersion you probably obtained the diagonalized Hamiltonian in the momentum space
$$
H=\sum_\mathbf{k}\left[c^{\dagger}_{\mathbf{k}A},c^{\dagger}_{\mathbf{k}B}\right]\left[\begin{array}{cc}0 & \Delta(\mathbf{k})\\ \Delta^{\dagger}(\mathbf{k}) &0\end{array}\right]\left[\begin{array}{c}c_{\mathbf{k}A} \\ c_{\mathbf{k}B}\end{array}\right].
$$
If you you chose your $x$ axis along the zigzag direction (arXiv:1004.3396), the two nonequivalent Dirac valleys are $\mathbf{K}_\kappa=\left(\kappa\frac{4\pi}{3\sqrt{3}a},0\right)$, $\kappa=\pm1$ and $\mathbf{K}_{-1}=\mathbf{K}^{\prime}$, where $a$ is the C-C distance. Then $\Delta(\mathbf{k})=-t\left(1+e^{-i\mathbf{k}\cdot\mathbf{a}_1}+e^{-i\mathbf{k}\cdot\mathbf{a}_2}\right)$, where $t$ is the hopping term, and $\mathbf{a}_1=\left(\sqrt{3}a/2,3a/2\right)$ and $\mathbf{a}_2=\left(-\sqrt{3}a/2,3a/2\right)$ are the lattice vectors.
Taylor expanding $\Delta(\mathbf{k})$ up to linear terms around those two points you obtain
$$
\Delta(\mathbf{k})=\kappa\frac{3ta}{2}q_x-i\frac{3ta}{2}q_y
$$
where $\mathbf{q}$ is the displacement momenta from the $\mathbf{K}_\kappa$ point. Promoting these displacement momenta to operators you obtain the Hamiltonian
$$
H=\hbar v_F\left[\begin{array}{cccc}0 & q_x-iq_y & 0 & 0\\q_x+iq_y & 0 & 0 & 0\\0 & 0 & 0 & -q_x-iq_y\\0 & 0 & -q_x+iq_y & 0\end{array}\right]
$$
where $v_F=\frac{3ta}{2\hbar}$ is the Fermi velocity. This is in $\left[\Psi_{A\mathbf{K}},\Psi_{B\mathbf{K}},\Psi_{A\mathbf{K}^{\prime}},\Psi_{B\mathbf{K}^{\prime}}\right]^T$ basis, if you rearrange your basis as $\left[\Psi_{A\mathbf{K}},\Psi_{B\mathbf{K}},\Psi_{B\mathbf{K}^{\prime}},\Psi_{A\mathbf{K}^{\prime}}\right]^T$ you get the compact form
$$
H=\hbar v_F\tau_z\otimes\boldsymbol{\sigma}\cdot\mathbf{k}
$$
where $\tau_z$ acts in the valley space. This is similar to the Dirac-Weyl equation for relativistic massless particles, where instead of $v_F$ you get the speed of light
$$
H=\pm\hbar c\boldsymbol{\sigma}\cdot\mathbf{k}
$$
where $+$ denotes right-handed antineutrions, and $-$ denotes left-handed neutrions. The differences are that $\boldsymbol{\sigma}=\left(\sigma_x,\sigma_y\right)$ for graphene acts in pseudospin space and $\boldsymbol{\sigma}=\left(\sigma_x,\sigma_y,\sigma_z\right)$ for neutrinos acts in real spin space.
Brillioun zone (BZ) generally refers to a domain in the space of $k$-vectors, and in that sense are geometric. Band generally refers to energy levels. The concepts are closely related, but they are different, so the glib statement that a BZ is a band is not quite correct, but most people understand what is meant. Every point in a BZ maps to one or more bands.
A dispersion relation (or "energy curve") maps points in the domain of $k$ space to the range of energies.
In the diagram you present, the shaded areas indicate regions of $k$ space that map to energies that are occupied. The unshaded regions map to energies that are unoccupied. Two bands are shown in the example in two different representation schemes: the two on the left are in the extended zone scheme, and the two on the right are in the reduced zone scheme. Bands are discussed in the context of the reduced zone scheme. In that scheme, the set of all energies in the range mapped to by $k$ vectors in the domain (the BZ, the square) constitutes a band. Note that each point in $k$ space maps to two energies, that is, to states in two bands.
Best Answer
The relationship of free electron to tight binding is understood via the Kronig-Penney model. In the Kronig-Penney model a series of quantum wells (particle-in-a-box) are separated by somewhat low walls which allow tunneling between the wells. In the free electron model we start by ignoring the walls and just "folding back" the parabolic energy vs. wave vector relationship and then using the potential of the walls as a perturbation. In the tight binding model we look at solutions to the states in the individual quantum wells and then see how they are modified by interacting with their neighbors. So, from a free electron view point, the lowest state has $k=0$ and would model an $s$ orital. The next state up is just a single wave cycle, so it looks like a $p$ orbital, etc... Within either model you never exhaust the possibilities, you just go higher and higher in energy with more nodes. When the height of the walls is low relative to the energy, the states look like free electrons and when the walls are high they look like the solutions of stand-alone quantum wells (and the band structure is relatively "flat"). For intermediate cases the solutions seamlessly transition betweeen the two models.