Now I am left wondering why does the heat become lost as if travels slightly.
It is not lost. It is spread more out.
If you stand so close to the heat source that you are hit by, say 1/10 of it's radiation (1/10 of all photons sent out hit you), then when standing further away you are maybe only hit by 1/100.
The heat radiation sent from the source like the sun spreads out with distance. The same amount of energy every second is spread out at a larger and larger sphere as it travels away from the source.
Surface area of such a sphere is $A=4 \pi r^2$. Doubling your distance to the campfire means:
$$A_2=4 \pi r_2^2=4 \pi (2 r_1)^2=4 *4 \pi r_1^2=4 A_1$$
The area that the radiation is spread over is four times as large for just the double distance.
This of course regards the sun with only space surrounding it. Since the campfire is place on ground, downwards radiation in that case will be smaller.
Solar constant
For your interest, the Suns' intensity on our planet Earth is called the solar constant (or solar coefficient) $S$. In units $\mathrm{W/m^2}$. This tells us how much radiation that reaches a square meter on Earth (or any other planet at the same distance from the Sun) every second. Mercury which is closer will have another solar constant.
To find the Suns' intensity at any distance, you will need to know how much energy is generated per second within the Sun, that is the power of the Sun $P$. This energy will be spread out while travelling:
$$P=S*A=4 S \pi r^2$$
where both intensity $S$ and area $A$ are for a specific sphere at a specific distance. The Suns' own intensity at its surface is then found by insertings the Suns' own radius and isolate S.
Also, our Sun can be viewed as a socalled blackbody. That is, it emits radiation very efficiently. The Stefan–Boltzmann law of blackbody radiation then gives us the emissive power from the Suns' surface:
$$P=\sigma T^4 A= \sigma T^4 4 \pi r^2 $$
with $T$ being the surface temperature of the Sun (around $5800^\mathrm{o C}$ if I remember correctly).
Campfire
Regarding the campfire as pointed out in the comments, convection might be considerable if you are standing very close to the campfire or maybe even reaching over it.
By natural convection, heated air will flow upwards, and this will carry a lot of energy that way. The radiation itself is negligible at that position.
Walking slightly further away might remove the convection effect entirely. This will feel like a huge decrease in heating. Adding a little windy weather, you might not feel any wind while being in "equilibrium" in the convection zone near the campfire. But walking two steps away can have a large cooling effect now that forced convective cooling from the wind acts as well.
Best Answer
Atoms and molecules are coupled with the electromagnetic field. The electrons of the atom and molecule are accelerating and then it will emit electromagnetic radiation, making the system losing energy.
Classical mechanics will predict a total collapse of the atom. Turns out that Quantum Mechanics saves the day, introducing the concept of discrete spectrum and ground state, and the ground state will be a stable state for the system.
Quantum mechanics can describe the process of radiation of atoms and molecules, see this for more details of how this works.
Now, in general, systems are stable when they minimize the potential energy because there is always a force pushing the system towards the direction that minimizes the potential energy, i.e. the force $f_k$ applied on a generalized coordinate $q_k$ is equal $$ f_k=-\frac{\partial U}{\partial q_k} $$
if the system is conservative. If the system is perturbed at a local minimum of $U$, this force will always try to bring the system back to the minimum, and then will be stable if the perturbations are sufficient small.
Atoms and molecules are stable under sufficient small perturbation for other reasons. They are at the ground state, and any perturbation of the system will be reverted into electromagnetic radiation and sent to infinity.