I am trying to calculate the exact delta-v from various existing upper stages with a given payload in a vaccum and in orbit, say LEO. I understand that according to the Tsiolkovsky rocket equation:
$$\Delta v = \mathrm{I_{sp}} \ g_0 \ \ln(m_0 / m_f)$$
I can find all the various variables for my craft except $g_0$ that I don't fully understand. Is it always fixed to 9,806? Is it related to TWR (that I can find as well)? Is it always 1?
$g_0$ is indeed an idealized "surface gravity" for Earth.1 As arbitrary and nonsensical as that is for many applications, you may be better off just directly using the effective exhaust velocity:
$$\Delta v=v_{e} \ln \left(\frac{m_{0}}{m_{f}}\right)$$
1From Wikipedia's Standard gravity:
The standard acceleration due to gravity (or standard acceleration of free fall), sometimes abbreviated as standard gravity, usually denoted by $ɡ_0$ or $ɡ_n$, is the nominal gravitational acceleration of an object in a vacuum near the surface of the Earth. It is defined by standard as 9.80665 m/s2</sup=> (about 32.17405 ft/s2). This value was established by the 3rd CGPM (1901, CR 70) and used to define the standard weight of an object as the product of its mass and this nominal acceleration.
For the second part of your question, local acceleration does matter for the thrust-to-weight ratio.
Isp is exhaust velocity where $g_0$ is divided out by arbitrary convention, your rocket engine performing the same in any gravitational environment. I agree with Christopher James Huff that just formulating the rocket equation in terms of exhaust velocity directly is less confusing.
But TWR is a ratio. The ratio between one force, the thrust, and another force, the weight of the spacecraft (or engine). Conveniently, it makes 1 the limit for when a rocket can get off the ground.
But only the thrust of the two is constant, as an equal amount of mass has different weights on different celestial bodies. So your TWR is 6x higher on the Moon than on the surface of the Earth. And not really well defined in freefall.
