There's a few questions in there, so I'll address them one by one.
What does A x B : C Mean?
Read this as A instances of a B number of inputs to C number of outputs. What you are looking for, if I understand correctly, is a 16x2:1 or a 32x2:1 chip. If C is more than 1, then your chip is significantly more complicated - you would no longer be selecting one input and connecting it to the output. Which leads well into the next sub-question -
Why are there $200+ chips for this simple function?
The specific part you linked is a 1x32:16 wide-bandwidth, DC coupled, buffered, video MUX, which can select any of it's 32 video inputs and output them simultaneously on it's 16 buffered outputs, with a gain of 1-2x. You could sorta think of it as a 16x32:1 with a lot of features. It's got quite a bit more inside than just CMOS switches. It isn't really designed for your function, which is...
How do I connect two memories to a master CPU
The most common method for hooking up multiple memory chips to a driver/controller/cpu is to use a tri-state bus. The address lines drive both chips, and the data bus is shared between all chips. Both chips should have a pin like "output enable", which can be controlled by the CPU. I found this article discussing memory buses at a rudimentary level - it has descriptive images. See Figure 8 for the gist of what I think you want. It is the simplest way of hooking things up, and the way I would recommend if the chips support it.
How would I make one?
Well, I think you were on the right track. CPU buses can be bi-directional, so intercepting the right output enable signal may be risky. The part you were probably looking for was a digital switch, something like this 16x2:1 FET mux. This is the cheapest one at $1.75 each. Wide bidirectional buses are best handled by ICs.
Note
I would check with the maker of your CPU to look for app notes and reference designs regarding memory buses. That will be the easiest way to see if you're on track.
What you are trying to do is a bit more complicated than you seem to realize. For general unknown I/O pins, it looks like you will need two microcontroller pins. One pin will be a output with maybe 5 kΩ in series. The other pin is a input connected directly to the test chip pin. This allows you to drive the pin high and low, and to leave it undriven. But in all cases you can see what the digital level is with the other microcontroller pin. With this setup, you can detect all the possible digital cases of the pin under test driving high, driving low, or high impedance.
If you stipulate that this test setup is only for digital logic chips and only at the same power voltage as the micrcontroller, then there is no need for analog.
Power and ground pins are another matter, however. These pins need to be tied solidly high or low during operation. In the arbitrary case you will need both low and high side switches on every pin since any of them could be power or ground. This gets messy fast.
I don't want to rain on your idea for a personal project, but frankly this strikes me as quite hard to get right and not much use if you do. 74xxx discrete logic is rarely used nowadays, and when I do need to identify chips I don't find reading the number off the chip all that bad.
How about a capacitor meter? Those little suckers aren't labeled anymore.
Best Answer
I think you're misunderstanding how the switch operates.
simulate this circuit – Schematic created using CircuitLab
Figure 1. Each pole of the switch consists of a common wiper contact, 'C', which can touch one of the other contacts, 1, 2 or 3, at a time. The switch is shown in position 1.
I think you're ON-OFF-ON requirement can be satisfied as shown on the right using only one half of the switch.
After details of switch added:
simulate this circuit
Figure 2. Does this arrangement using one pole of the switch satisfy your requirments?
Table 1. Truth table assuming pullup resistors on GPIO.