You're substantially correct on everything you've mentioned. Bigger cable has lower losses.
Low loss is important in two areas
1) Noise
The attenuation of a feeder is what adds Johnson noise corresponding to its temperature onto the signal. A feeder of near zero length has near zero attenuation and so near zero noise figure.
Up to a meter or several (depending on frequency), the noise figure of a typical cable tends to be dominated by the noise figure of the input amplifier you are using, even cables of pencil diameter (you can get really thin cables, sub-mm even, and in these you do have to worry about meter lengths).
To get signals down off your roof into the lab, any feasible cable will be so lossy, even unusually thick ones, that the solution is almost always an LNA on the roof, straight after the antenna.
That's why do tend not to see really fat cables in labs, they're not needed for short hops, they're not sufficient for long drags.
b) High power handling
In a transmitter station, you tend to have the amplifier in the building, and the antenna 'out there' somewhere. Putting the amplifier 'out there' as well is usually not an option, so here you do have fat cables, as fat as possible given that they have to remain TEM, without moding. That means <3.5mm for 26GHz, <350mm for 260MHz etc.
The impedance of the cable also matters, as well as the size. Have a look at this cable manufacturer's tutorial on why we have different cable impedances, so 75\$\Omega\$ for lowest loss, and 50\$\Omega\$ as a compromise that has settled itself in as a standard.
Johnson noise is just the statistical variation of the voltage across a resistor that is due to the random motion of the charges within it.
If you work out the numbers, the amount of power that this represents is very very tiny — even when measured over a 1 GHz bandwidth, we're just talking about 4 picowatts (-84 dBm). If you wanted to produce 3.3V from this, you'd only get about 1 picoamp, assuming you could find a diode whose leakage is significantly less than that.
Note that in the case of two resistors with the same value, the Johnson noise power will divide equally between them, with half dissipated in the source (hot) resistor and half dissipated in the load (cold) resistor. But also note that any real circuit will have nonzero values of parasitic inductance and capacitance, and these will serve to limit the bandwidth of the noise transfer, causing more of the power to stay in the source and less to be transferred to the load.
Best Answer
To first of all answer the title of this question, modeling noise is not a waste of time as long as the device models are reasonably good and include the dominant noise sources. 1/f noise, in particular, becomes insignificant above a certain frequency, and in many applications is not detectable beyond other noise sources (never mind not being a significant noise source).
1/f noise is not (inherently) modeled in LTSpice (or SPICE in general), and depends on the particular resistor.
To model excess noise, you could try the technique illustrated below. In this file, you can see the 1/f noise at node "noise". To use it in a circuit, place E1 in series with the resistor you want to add 1/f noise to.
In doing this, you will need to know the resistor's noise index in uV/V. Since excess noise is proportional to the resistor's DC bias, you will also need to know the DC bias of the resistor. This technique will work if the resistor's DC bias is fairly constant. Edit the VALUE2 field of E1 (laplace=1*659/sqrt(s/6.2832)), and change the 1 to the product of the DC bias and the noise index.