Well, heat isn't translational energy specifically, or any energy specifically. It encompasses all the energetic degrees of freedom of a system or molecule. Or at least as long as the various degrees of freedom have some thermodynamic connection to each other. It's just that electronic states (for instance) have such large energy differences that the thermal energy doesn't excite them to any significant extent at 'ordinary' temperatures.
Now, as for what happens during bond formation, the way things work is that the electronic and nuclear kinetics are largely decoupled (the Born-Oppenheimer approximation). So to a large extent there's no _direct_ transfer of kinetic energy between the electrons and nuclei. The reason for this is that the nuclei are very very slow compared to the rate of movement of the electrons, so the electrons can be assumed to adjust instantaneously to any movements in the nuclei. (from the nuclear point-of-view) That has the consequence that there's no kinetic energy transfer.
So, if we look at things through that approximation (and it's a very good one), what happens during chemical bond formation is that the electronic energy gets lower; both in terms of kinetic and potential energy (so it's not purely electrostatic). But from the nuclear point of view, only the potential changes in our approximation, the attraction the nuclei feel is purely electrostatic here. What happens when the atoms get closer and bond is that the electronic energy drops and forms a region of electron density between the atoms, so the electrostatic potential changes; the atoms aren't free particles anymore, but now have a potential-energy well dependent on the interatomic distance that looks http://www.doitpoms.ac.uk/tlplib/stiffness-of-rubber/images/image01.gif" . They're bound.
Now in quantum mechanics, bound particles have discrete energy levels, and that goes for the atoms too, so that potential well has a bunch of energy levels, which are the vibrational energy levels - since that's what the bond stretching motion is. If the atoms just bound to each other, the nuclei still have their kinetic energy from when they were free particles (> 0 energy), meaning they're energetically still at the 'top' of this well. Indeed, a reacting pair of atoms has a probability of rebounding completely and not forming a bond. But more often, the nuclei "fall into" the well, losing vibrational energy to their surroundings. And that's what we regard as heat here.
But note that the total energy doesn't change of course. So why wasn't it 'heat' before? The answer to that is that if you have a chemical by itself, it can't convert that chemical potential energy into the other forms. So that particular 'chemical reaction' degree of freedom isn't part of its temperature. Once you mix two chemicals and they can start getting at that potential energy of course, then that energy is now available to be converted into rotational and vibrational and translational and the other degrees of freedom. So it will react and convert the chemical potential energy into the other forms until its in thermodynamic equilibrium. Just as with the electronic states, if the reaction energy is high, the thing reacts completely. If the chemical reaction energy is low, then you have a reaction equilibrium as part of the thermodynamic equilibrium. In other words, the concentration of reactant and product is part of the 'heat' of the system just as much as the distribution of speeds among the molecules and such.
As a short summary: Your description is mostly right. When two atoms get close and bond, the interactions of their electrons create a negative potential between the two nuclei, causing them to increase their kinetic energy, which can then be dissipated. But it's all heat. You could say the 'heat' was created the instant the two chemicals were mixed and thus able to react, and that the heat you feel is actually just 'thermal conduction' from chemical energy to the other forms.