Chemical bonding and hybridization of organic compounds
You talk about hybridisation but you haven't actually looked at any hybrid orbitals in your description. Since the geometry around the carbons is trigonal planar, each carbon hybridises its $\ce{2s}$, $\ce{2p}_x$ and $\ce{2p}_y$ orbitals to form three $\ce{sp^2}$ orbitals which lie in the $xy$ plane at $120^\circ$ angles to each other. One $\ce{sp^2}$ orbital on each carbon points towards the other carbon and so these orbitals overlap to form a $\ce{C-C}$ $\sigma$ bond. The other two $\ce{sp^2}$ orbitals on each carbon overlap with the $\ce{1s}$ orbitals on the hydrogens forming $\ce{C-H}$ $\unicode[Times]{x3C3}$ bonds. The remaining unhybridised $\ce{2p}_z$ orbitals on each carbon overlap above and below the plane of the $\unicode[Times]{x3C3}$ bonds to form a $\ce{C-C}$ $\unicode[Times]{x3C0}$ bond.

While $\ce{sp^2}$ orbitals still have two lobes, one lobe is substantially larger than the other and only that takes part in bonding. Note that this is the usual case: even in molecules that solely use p-orbitals, such as chlorine, only one lobe of a p-orbital will form a $\unicode[Times]{x3C3}$ bond; the other of the same orbital will not.

This link provides a nice visualisation of the different orbitals involved in ethene bonding.