Why does the antibonding orbital has higher energy (than the bonding orbital) if the Coulomb repulsion is lesser? Depending on its strength, the attractive double dirac delta potential shown below

can support two bound states. They are called the bonding and the antibonding orbitals as shown in the figure below

It turns out that the antibonding orbital has higher energy than the bonding orbitals.
But from the figure above, it is cleat that the electronic charge density piles up more in between in case of bonding orbital and less in case of antibonding orbital. Does this not mean that the Coulomb repulsion between electrons be more in case of bonding orbital configuration than the antibonding case. If so, doesn't that mean the energy of the bonding orbital should be more than the antibonding orbital?
 A: A single electron does not experience Coulomb repulsion with itself. The potential and wavefunctions you describe involve only one-electron, so the state energy cannot be associated (even as an approximation) with the Coulomb repulsion between two electrons.
On the other hand, the double delta potential can be seen as a very crude approximation of the Coulomb potential of two positive nuclei. The fact that for the bonding state, the electronic charge density piles up more in between  merely reflects that the electron is more likely to be found between the two nuclei, justifying the name "bondig". The opposite holds for the anti-bonding state.
For the sake of argument, let us add another electron to your model and ask what is the ground state of the two-electron systems?
Without explicitly stating other assumptions, you model again does not imply any interaction between the electrons beyond Pauli exclusion principle. If these electrons were of opposite spins, they would both occupy the lower-energy bonding state. If they were of the same spin, one would be in the bonding state and the other in the anti-bonding one.
Finally, let us ask if we have two electrons of opposite spins in the bonding state and we then "turn on" the Coulomb repulsion between the two, what would happen?
It depends on whether the Coulomb energy between two bonding states  is larger than the energy increase resulting from moving one or both electrons to the anti-bonding state. If it is not, the electrons stay in the bonding state and you get a covalent bond.
A: The $\delta$ potentials may be the main reason why this is a problem. In reality, the nuclei have $1/r$ potentials. The potential for a single electron is lower between the nuclei than on either side.
When the electron wave function has a node between the nuclei, the negative charge density is low. The positive nuclei also feel each other's $1/r$ potential and will then repel each other.
The equipotential surface can then be dissociative.
A: As @Quantum-Collapse has noted in their answer, there is some confusion here between bonding and anti-bonding states which refer to the one electron problem, and the singlet and triplet states for two electrons. The former problem arises, e.g., when discussing the covalent bonds, in which case the bonding/symmetric state has lower energy, whereas the latter arises, e.g., when discussing the Helium atom where  the synglet state, whose orbital wave function is symmetric in respect to exchanging the electrions, is the ground state. Note that the symmetries are different here. The two problems get mixed when discussing the covalent bond in the hydrogen molecule.
To get closer to the question: the node theorem (see, e.g., here and here, but also in many textbooks) states that the ground state of a one-dimensional Schrödinger equation does not have nodes, whereas the n-th state has $n-1$ nodes. (Note that this is strictly true only in one dimension). The physical explanation is that nodes correspond to zeros of the probability density, i.e., the electron is confined to smaller regions in space, which means lower uncertainty in position and higehr uncertainty in momentum, hence the higher energy. This is admittedly a very hand-waving explanation, but the mathematical result is rigorous.
A: There is a confusion between single electron and multi-electron states here.
First consider the single electron states. The bonding orbital has the lowest energy because it allows the electron to have the lowest potential energy. This is true as long as the bound orbital of V1, $\psi_1$, overlaps with V2 and vice versa. For the bonding orbital $$\Psi_B=\frac{\psi_1+\psi_2}{\sqrt{2+2S}}$$ the two orbitals constructively interfere at the potential positions, which gives it the lowest potential energy. For the antibonding orbital $$\Psi_A = \frac{\psi_1-\psi_2}{\sqrt{2-2S}}$$
the two orbitals interfere destructively. Here $S=\langle \psi_1 | \psi_2 \rangle$. The electron self interacts but this energy is independent of the orbital in the nonrelativistic limit and is considered absorbed in its rest energy.
For two electron states one can construct the bound state $$\Psi_0=|\Psi_B(1) \overline\Psi_B(2)|~,$$ where || indicates a Slater determinant. This state has the highest electron-electron repulsion but this is more than compensated by the lowest potential energy, which makes it the ground state. Note that $\Psi_0$ will have some admixture of $|\Psi_A(1) \overline\Psi_A(2)|$. The triplet
$$\Psi_1 =\frac{|\Psi_B(1) \Psi_A(2)|}{\sqrt{1-S}}$$
is the first excited state, followed by the corresponding singlet $$\Psi_2 =\frac{|\Psi_B(1) \overline\Psi_A(2) - \overline \Psi_B(1) \Psi_A(2)|}{\sqrt{2+2S}} ~.$$
