"You can't use lenses and mirrors to make something hotter than the surface of the light source itself."
From the above you should be able to see that this is complete BS. For a real world example, a semiconductor laser can burn a hole in steel. The melting point of gallium arsenide (or whatever chip material) is a lot less than that of steel. The steel melts. The chip doesn't.
You're very nearly correct in your understanding of this subject, but it is imperfect --
The radio example works because the radio's front end has an effective temperature below that of the signals it's intended to receive. Which depends on what frequency band it's designed for, because background noise varies, but let's say VHF or UHF where we can ignore atmospheric noise and consider the background being faint (except for the Sun) astronomic noise sources.
The radio consumes power, and consequently is able to produce nonreciprocal elements. One consequence of this is, the effective noise temperature of the front end can be much smaller than the ambient temperature of the instrument as a whole. Indeed, adequate to resolve those distant astronomic noise sources, for some radios (although it's noteworthy that some of those radios are, in fact, cryogenically cooled, for even better performance; I don't know when that's required as far as effective noise temp).
It's noteworthy that all the elements up to the radio front end (transmission medium, antenna, transmission lines) are (usually / preferably) low loss. That is, they do not contribute much of their own temperature to the noise level received. The optical equivalent of this is simply a mirror: a metal mirror can be quite hot, incandescent even, but it is still shiny. Its emission only dominates over the reflected signal when the reflected signal is below the emissivity of the mirror.
Or for transmissive media, same thing, the bulk glow doesn't necessarily dominate. Hot glass looks really freaking cool: it does glow, but not much, because its emissivity in visible light is very low; meanwhile, it remains clear and colorless, so you can see through it, but also see its own glow through it.
So too, the antenna needs to be low loss, so that it faithfully couples between free field and transmission line, without adding its own noise to the system. In effect, the antenna is a lens, and the transmission line is a fiber optic, shiny and reflective and clear, without adding its own noise to the system. (Of course, a sufficiently bad, and long, cable will have appreciable losses, and introduce its noise and temperature to the system. Such is the case with any medium.)
Conversely, the laser is a special, active device. I'm not too certain of the V-I vs. illumination characteristics of a solid state laser, but at the very least, as a PN junction, it must act like a solar cell. Which responds to incoherent, wideband radiation, not just sharply tuned laser light. So already we can see there is some nonreciprocity here. (Well, anything inside a resonator will respond mainly to signals passed by that resonator -- but the point is, the response will still be wider than the laser emission line(s) are.)
The laser emission is extremely intense (not just narrowband and coherent, but very parallel as well), and this occurs due to a feedback process. The equivalent temperature is very high indeed: to have such high emission at a narrow band, you'd need an incredibly hot source (like, Sun's core hot), sent through an incredibly narrow bandpass filter.
Why does it melt metal, but not itself? Well, reciprocity is one thing; lasers can indeed be damaged by very reflective targets (poorly matched loads -- so too can radio transmitters!). As long as most of that energy is leaving, it's fine.
You might just as well ask why an induction heater can melt metal, when its energy is very low (fractional MHz, so, whatever that is, photons of nanoelectronvolts?). Clearly its effective temperature must be incredible (and the bandpass on that noise, equally strong), and to get visible light emission somehow or another implies one heck of a multi-photon mixing/upconversion process. Well, something like that. Physics gonna physic. Suffice it to say, a quantum description of a macroscopic, classical system is a rather impractical approach, and since it happens, we can conclude that yes, somehow or another, such a mechanism exists.
So, suffice it to say: classical physics is adequate here, and active devices not in thermodynamic equilibrium, need not obey the laws of thermodynamics on a local basis?
Tim