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.
The best way to get a proper understanding is to write down the truth table completely. You have 3 inputs (A, B, S), so this will give you 2\$^3\$ = 8 combinations:
S A B | S' C D | Z
--------+---------+--
0 0 0 | 1 0 0 | 0
0 0 1 | 1 0 0 | 0
0 1 0 | 1 1 0 | 1
0 1 1 | 1 1 0 | 1
1 0 0 | 0 0 0 | 0
1 0 1 | 0 0 1 | 1
1 1 0 | 0 0 0 | 0
1 1 1 | 0 0 1 | 1
It's often useful to add intermediate results to make things more clear. I added a term \$C = (A \land S')\$ and \$D = (B \land S)\$. Now it should be clear that \$Z = (C \lor D)\$.
Best Answer
As long as you allow ground as an uncounted line, you can do it a bit more simply than Asmyldof suggests. Not only that, you can do it for as many channels as you like.
simulate this circuit – Schematic created using CircuitLab
The counters are synchronous reset, and the carry is active when the counter is at max carry. The polarities of the carry and reset must be the same. 74HC163 is a type IC, except that the carry and reset are of opposite polarities, so an inverter must be used.
There are only 3 signals required: clock, reset and data.
The counter size can be more or less indefinitely extended.