There isn't a single value for the lifetime of a horizontal branch star. The lifetime depends upon star mass ($M$), helium core mass ($M_c$) or helium fraction ($Y$), and metalicity ($Z$), where the masses are in units of solar mass.
According to Iben's POST MAIN SEQUENCE EVOLUTION OF SINGLE STARS:
log (lifetime in units of $10^7$ years) $ = 0.74 - 2.2(M_c- 0.5) +log(f)$
$= 0.795+0.506( Y-0.3) + 0.022(log(Z) + 3)+0.077(M -0.8) + log(f)$
where $f$ is a convection factor that is approximately $2$.
In one example of Iben, $Y=0.3$, $Z=0.001$
In another $Y=0.3$, $Z=0.0001$
So $10^8$ years seems like a good estimate for a star of 0.8 solar masses, while $10^7$ years seems too low.
Short answer: gravitational potential energy is converted into heat.
Let's look at the Sun as an example. Its mass is $M_\odot = 2.0\times10^{30}\ \mathrm{kg}$ and its radius is $R_\odot = 7.0\times10^8\ \mathrm{m}$. If its density were uniform, its gravitational binding energy would be
$$ U_{\odot,\,\text{uniform}} = -\frac{3GM_\odot^2}{5R_\odot} = -2.3\times10^{41}\ \mathrm{J}. $$
In fact the Sun's mass is centrally concentrated, so $U_{\odot,\,\text{actual}} < U_{\odot,\,\text{uniform}}$.
Where did the Sun come from? Something like a giant molecular cloud with a density of $2\times10^{-15}\ \mathrm{kg/m^3}$. The mass of the Sun would thus have been extended over something like a sphere of radius $6\times10^{14}\ \mathrm{m}$, for a gravitational binding energy of
$$ U_\text{cloud} = -3\times10^{35}\ \mathrm{J}, $$
which is negligible in comparison with $U_\odot$.
All of the $2.3\times10^{41}\ \mathrm{J}$ had to go somewhere, and the only place to dump energy is into heat. The gas particles gain velocity as they fall into the potential well, but they don't lose that velocity because they never climb back out of the well.
Not worrying about whether the heating is isobaric or isochoric or somewhere in between, the heat capacity of monatomic gas is about twice the ideal gas constant, or $8.3\times10^3\ \mathrm{J\,K^{-1}\,kg^{-1}}$. At this amount, in order to heat all of $M_\odot$ by the average temperature of the Sun (say $10^7\ \mathrm{K}$, somewhere between the core and surface temperatures), you would need about $1.7\times10^{41}\ \mathrm{J}$ of energy. There is enough energy released by gravitational collapse to heat the Sun to its current temperature. You can do a more detailed analysis taking into account how much cooling occurs during the collapse, but the steep temperature dependence of the Stephan-Boltzmann law makes it difficult to lose heat to space until the object is already hot. I'm also neglecting a factor of $2$ that comes from splitting the energy between heating the gas and compressing it.
Once the material is this hot, it simply glows like any blackbody emitter. The energy lost to space is replenished by nuclear fusion in the core. In fact, fusion acts as a regulator: too much of it and the star expands and cools, slowing down fusion; too little and the star collapses further, heating up more and increasing the fusion rate.
In summary, gravitational collapse provides the initial energy to heat a star. As it uses up this energy source, it begins to tap into fusion. Ultimately it reaches an equilibrium where the energy produced by fusion is balanced by the energy radiated into space.
Best Answer
An ADS search for "star formation" turns up about 142,000 articles with "star formation" in the title or abstract. The first article is a 43 page review paper of Star Formation in Galaxies in the Hubble Sequence, written by Robert Kennicutt, Jr, one of the leaders of the field. He never defines anything else to mean star formation and one of the "key words" used for tagging the article is "star formation."
I think it's safe to say, "star formation" is the technical term that everyone uses. Note, though, that as an extension to Jim's comment, star formation is generally used in the context of a region of space that is currently forming stars, not for a single star that is being formed.