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.
updated calculations - based on neutrino energy escaping and vapor inhalation risk
Your math is close but not quite right.
First - the number of tritium atoms.
There are 1000/(16+3+3) = 45 moles (as you said)
This means there are 45*2*$N_A$ = $5.5 \cdot 10^{25}$ atoms of Tritium
Now the half life is 12.3 years or 4500 days, that is $3.9\cdot 10^8 $s.
This means the 1/e time $\tau=t_{1/2}/\ln{2} = 5.6\cdot 10^8 \mathrm{s}$
The number of decays per second is $\frac{5.5\cdot 10^{25}}{5.6\cdot 10^8} = 9.8\cdot 10^{16} s^{-1}$
The energy given off in one decay is 19 keV (source: LBL) - but the mean energy imparted to the electron is less: only 5.7 keV (Wikipedia. The remainder goes to the anti-neutrino that is also produced in the decay. Since the interaction cross section between the neutrino and the water is extremely small, that energy can be considered "lost" in terms of heating the water.
This means the total energy deposited in the water (all of it, since the beta decay has a very short range) is about 90 W
Clearly this is a significant source of heating - about 22 calories per second, so it will heat a liter of water by one degree C in about 50 seconds. In an insulated container, liquid that started at room temperature would boil in about a little over an hour. During that time, it would produce about 14 ml of $^3He$ gas as well (0.6 mmol).
This liquid would probably not glow - the blue glow usually associated with radioactivity is due to Cerenkov radiation (roughly the optical equivalent of a sonic boom) and that requires particles to travel faster than the speed of light in the medium.
An electron with 19 keV is not really relativistic (rest mass 511 keV), so the speed is given near enough by
$$v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2\cdot 19\cdot 10^3 \cdot 1.6 \cdot 10^{-19}}{9.1\cdot 10^{-31}}} \approx 8\cdot 10^7 \mathrm{m/s}$$
This is sufficiently below the Cerenkov limit (which for water with a refractive index of about 1.3 would be about $2\cdot 10^8 \mathrm{m/s}$ - so there will be no blue glow.
But just because it doesn't glow does not mean it's not dangerous.
According to the CDC, a skin dose of 550 Gy would require amputation. Now with half the water near the skin radiating "away" from the skin, we expect the deposited dose per unit mass in the skin to be half of that deposited in the water; this means the dose rate is 45 Gy per second. You would reach the "amputate everything" dose in about 12 seconds. But because the beta has such low energy, it is said to be absorbed in the "dead upper layer of the skin". This might protect you... If it weren't for the fact that if you have tritiated liquid, you have tritiated vapor. Inhalation of that vapor will be really bad for you.
How bad? Assuming that the swimming pool is a comfortable 22 °C, the saturated vapor pressure of water at that temperature is about 20 Torr or 260 Pa. At 60 % relative humidity (not uncommon near a pool) that would be 150 Pa, or about 1/600th of the air. Assuming that all that inhaled tritium is exchanged with the water in the body, and that an average human exchanges 1 m3 of air with the environment per hour (source), this means 1.6 liter of tritium gas per hour - about 70 mmol. This would deposit 0.07 Gy per hour into the body - where there is nothing to protect you. It would take a bit longer to kill you... but kill you it would. Of course swallowing a bit of pool water would really speed up the process, as would any cuts or abrasions. And did you ever notice how crinkly your skin gets when you spend too much time in the water? That's water penetrating the skin. I don't have a good way to estimate that - the internet is full of stories on the subject, but I found no hard data. I suspect that it will be a significant source of tritium entering the body if you swim in your pool.
So even a very short exposure is likely to be most unpleasant. I recommend against using this as a method to heat your swimming pool.
As an aside, there are plenty of instances where Tritium is used as a good "permanent" light source for watch hands, sighting compasses, and other instruments. The short range of the beta particles and the practical half life make it quite good in this role; mixed with an efficient scintillator that turns the energy into visible light, you don't need a lot of tritium to light up a dial at night. You are allowed to use 25 mCi of tritium for this application without needing an NRC license.
Now 25 mCi is 925 MBq - million decays per second. Your liter of tritiated water exceeds this by about 9 billion; and it is in direct contact with the skin (instead of isolated between a few mm of glass which effectively stops the beta radiation).
This is why tritium watches are safe, and your swimming pool is not.
Best Answer
The reason being closer to a heat source makes you warmer is the inverse square law. Think of it this way: If you have a $1~\mathrm{m}^2$ piece of material facing the Sun and located at Mercury's orbit, it will be quite hot. What does the shadow of this square look like at Earth's orbit (about $2.5$ times further away than Mercury)? Well, it will be $2.5$ times bigger in both directions, covering about $6~\mathrm{m}^2$. So the same amount of power can be delivered either to $1~\mathrm{m}^2$ on Mercury or to $6~\mathrm{m}^2$ on Earth. Every square meter of Earth gets about $6$ times less Solar power than every square meter on Mercury. The light is not losing energy to the surrounding medium, even if the medium exists.