To the OP: I don't want to go too far down this rabbit hole, but I'll try to give some kind of useful response. You obviously ran across the idea of "equivalence" as it relates to digital photography and are wondering if it applies to film photography, too. The short answer is that it does, but for folks who cut their teeth in the film era (of which I am one) it's kind of an alien concept. Besides a handful of scientists, very few of us thought in those terms before people started delving into the quantum physics and information theory that underlies digital imaging (and film imaging, too, since it underlies everything in existence).
But first, I'll say that I think you need to get a clearer grasp of equivalence theory, what it's good for (it has fairly limited usefulness) and of how light falling on an object is measured and described. A big reason why you got such a confusing mix of answers to your questions is that your terminology is not very clear. So then:
I know for camera sensors, for example a micro 4/3 sensor is less sensitive to light than a full frame (35mm equivalent) sensor. At the same aperture and shutter speed, a full frame sensor is capable of much better low light performance because it takes in more light.
The phrase "sensitive to light" and "takes in more light" are a tangled knot here. For any given image that you frame the same way (i.e. let's say you take a picture of a red 1957 Pontiac that fills the image frame), with the same exposure settings, a full frame sensor does capture more _total light_ simply because it has a bigger area. Another way to think about this is that the image of the Pontiac is larger when you capture it on a FF sensor. But the light _per unit area_ will be the same if the f-stop (a.k.a. 'f-ratio' or 'relative aperture') and shutter speed are the same. So, intensity of light (not a scientific term) at any given point on the sensor would be the same, but the total amount of light used to form the image would be greater on the FF sensor. If you painted the Pontiac equally bright with red paint to fill a 5x7 canvas and then again to fill a 16x20 canvas, the luminance of the red paint would appear the same on both paintings but you'd have to use a lot more paint to make the 16 x 20.
With some very small exceptions, it's the greater total amount of light that allows the larger sensor to capture a higher quality image ("better low light performance", as you called it, is only one of the ways it's better). It actually has almost nothing to do with the size of the pixels, as some people think. This has been proven definitively. The reasons are rooted in quantum physics and the quantum nature of light. There are some tutorial articles at DPReview that will demonstrate this if you doubt it.
How about medium format, is it more light sensitive due to having a bigger "sensor" or "contact area"? For example, a 75mm 4.5 medium format lens, does it take in the same light at 4.5 as a 75mm 4.5 35mm film camera lens? Or more light?
So the answer here is yes, the larger film area has the same benefits for film as the larger sensor area has for digital. And it's for the same reasons, all rooted in quantum physics. But film photographers never talked about it in this way, so if you come to a great forum like this and ask equivalence questions, it seems like you're speaking a foreign language. But it's not a foreign language; film photographers know it intuitively, as follows:
Let's say your end goal is an 8x12" print of our 1957 Pontiac. So you take two pictures; one where you fill the frame of a 35mm negative on ISO 100 film and shoot the picture, and a second where you fill the frame of a 6x9 cm negative on the same film. [Note that to get the same depth of field you have to shoot the 6x9 camera at a bit more than two stops narrower aperture. That means you have to use a shutter speed that's about two stops slower. No biggie because, for this example, our Pontiac is not moving.]
Then you make 8 x 12" prints from the negative. Every film photographer knows the print from the 6x9 negative will be technically better; we say that the 6x9 negative requires much less enlargement to reach an 8 x 12 size and that's why it's better. But another way to say that is that the image is bigger on the 6x9 negative; it was formed with more total light. (Quantum physics rears its head!)
Now, let's extend this idea to the realm of "better low light performance". Let's shoot the Pontiac on our 6x9 camera using ISO 400 film instead of the ISO 100 film we used in the 35mm camera. Now, when we use a narrower aperture to equalize the depth of field, we don't have to bump our shutter speed down (because our film is two stops faster). We can use the same shutter speed in both cameras. So if our Pontiac was moving, the two pictures would still be the same -- same depth of field, same motion freezing effect.
And if we make 8 x 12 prints from the negatives, now they are (roughly) the same quality. Yes, the 6x9 negative is bigger and is enlarged less, but the ISO 400 film is grainier than the ISO 100 film (plus other differences) we used in the 35mm camera. So note: in the 6 x 9 camera, the higher ISO film (better low light capability) produces the same quality 8 x 12 print as ISO 100 film in the 35mm camera.
In that last scenario, we've also equalized depth of field and shutter speed. Every parameter is the same, and we end up with equal quality prints (within the limits of this approximation). This is all the equivalence theory does: it tells you what settings/parameters are needed to produce the same result on different sensor/film formats. In this case, to produce the same 8 x 12 print, in the 6x9 camera we would use iso ISO 400 film, two stops narrower aperture (compared to the 35mm camera), same shutter speed; and in the 35mm camera, ISO 100 film, two stops wider aperture, and same shutter speed.
Now, before those who really know their math chime in and say so: the equivalence theory difference (derived from imaging area) between 35mm and 6x9 is not exactly two stops; it's a bit more than that, but I simplified it to two stops for the purposes of illustration.
Is there some sort of "equivalency" in the numbering or is there any difference in metering here to be aware of...
As everyone else pointed out, no. There is no equivalence math to do when you're taking pictures. Absolutely no difference in metering. The f-stop or f-ratio denotes the same thing regardless of the size of your sensor/film. What it denotes is amount of light per unit area, (or intensity, if you like that word better) but not "amount of light" if you define that term as total light used to form the image.
The two things to be aware of that relate to equivalence are (assuming the same framing of the subject): compared to 35mm FF, you will need narrower apertures on medium format to achieve the same depth of field (how much narrower depends on which medium format you are using -- i.e. 6x4.5, 6x6, 6x9 etc) and you can use a higher ISO film on the medium format camera and get the same print quality as you would with lower ISO 35mm film _at any given print size_ (how much higher ISO again depends on the which medium format you are using). This last fact was not a big part of film era thinking -- most people used medium format to get a better print (or scan) with the same low ISO film they might use in their 35mm camera. But some wedding photographers relied on it, even if they didn't know they were; they could use ISO 400 film to extend the range of their flashes and otherwise make event shooting easier than it would have been with ISO 100 film, yet still provide very good quality proofs and enlargements from their medium format negatives. Much better than they would have been able to provide using ISO 400 film in a 35mm camera. Again, in those days we attributed this to the bigger negative that needs less enlargement for any given print size; but you can express it in equivalence terms by saying it's due to the bigger image, formed by more total light.
I didn't do a very good job of not going down this rabbit hole. Oh well.
