I think what OP meant as a question is more elementary, i.e. in a course context.
In general, the reason why we think of a pigeonhole argument is because we think there is enough elements we need to choose from some lot for these elements to get something out of this lot. Let's just take an example.
Suppose you choose randomly points in an equilateral triangle with sides length $2$. Prove that wherever we take $5$ points inside this triangle, there exists at least a pair of points with distance between them less than $1$.
The key here is to see why does $5$ points is enough to know that two points will be close enough.
Suppose we took $3$ points first. To make sure they're as far as possible, let's put them at the vertices of the triangle. Now we want to add a fourth point. Well, if it's as far as possible from the four others, then it has to be in the middle. But now where do I put the fifth point?
Solution : If you divide the triangle of length $2$ into four triangles by tracing lines between the midpoints of the sides of the original triangle (i.e. by dividing it into four triangles of side length $1$), you realize that wherever I put the $5^{\text{th}}$ point, it has to lie in one of those triangles, and two points inside such a triangle have distance less than $1$. Here! We've found pigeonholes : the triangles of length $1$. If these are the pigeonholes, then we're trying to put $5$ points in $4$ pigeonholes, hence at least $2$ points are in the same triangle, so they have a distance less than $1$.
The idea behind my research of the solution was that I was trying to "find some optimal position" for the points, and then realized "there's not enough space for them". The part where you notice there's not enough room for all your things is where the pigeonhole principle usually does the trick.
To answer your second question, to choose the pigeonholes, you need to make them work for what you're looking for. For instance, here I wanted pigeonholes which ensured me that two points inside those regions gave me my result. Triangles did the trick so I just guessed them from looking at the problem long enough.
For another example, if you're looking at, say, trying to say that some set of numbers must contain a pair whose difference must be $n$, then it might be a good idea to pair up $k$ and $k+n$ together, since their difference is $n$. The idea here is simply to construct pigeonholes in such a way that when you have two things in your pigeonhole, you're able to solve your problem. Sometimes adding pigeonholes with less than two things in it might be pertinent though (to eliminate trivial cases), and in this case we know that two things cannot have been taken in this pigeonhole since it has only one thing in it.
Hope that helps,