Symmetry of an in-phase and an out-of-phase wavefunction "approaching" each other in an MO
Considering this diatomic molecule, the total wavefunction $\Psi$ is formed via a linear combination of the atomic orbitals $\psi_1$ and $\psi_2$ (which is an approximation, but not a too bad one).
$$ \Psi^+ = c_1 \psi_1 + c_2 \psi_2 $$
$$ \Psi^- = c_1 \psi_1 - c_2 \psi_2 $$

The (electronic) energy of the system can be described using the following (stationary) Schrödinger equation:
$$ H_\text{el} \Psi_\text{el} = E_\text{el} \Psi_\text{el}$$

So somehow this all factors into each other and in the end gives you the curves (for the potential energy of the system vs. internuclear distance) that you posted. Where is the symmetry? Well, it is still there, with respect to the center of one atom. If you imagine rotating the functions around the $y$ axis, you will get a symmetric image. And you're allowed to do this because space expands in all three directions. So in theory (if you follow through and expand the function to everywhere) you get a 3D image of the energy at every point around one atom.



But: The wavefunction in itself is not all that interesting to us, because there is no "real" physical information that can be directly extracted from it. But fret not, we can at least get out the electron probability density by complex-squaring it:
$$ \rho = |\Psi|^2 = \Psi \Psi^* $$

And this is where those nice small bubbles come into play that we see in your second figure. And we have again found the symmetry that you seek, because both linear combinations have $D_{\infty h}$ symmetry (I think... I'm not all too good with point groups and the such).