Here is an example from "Discrete and combinatorial mathematics an-applied introduction" by Grimaldi. My problem is that I have a hard time following along with the arguments.
Example: For $n\geq 1$ we want to tile a $2\times n$ chessboard using the two types of tiles shown in this Figure: 
Solution: To the left you see both of the tiles and an example for a $2\times 2$ chessboard to the right. Letting $a_n$ count the number of tilings, we find that $a_1=1$, since we can tile a $2\times 1$ chessboard (of one column) in only one way -- using two $1\times 1$ square tiles.
For $a_2$: Since the left tile takes up 3 "slots" and the single square just 1. So by rotating the left most piece would mean you could do it in 4 ways (and then always place the single piece in the empty spot. Or you could also just take 4 of the single square and place it in every square on the chessboard. This covers all the possibilities and $a_2=4+1=5$.
Finally for the $2\times 3$ chessboard there are 11 possible tilings (i) one that uses six $1\times 1$ square tiles; (ii) eight that use three $1\times 1$ square tiles and one of the larger tiles; and (iii) two that use two of the larger tiles. When $n\geq 4$ we consider the nth column of the $2\times n$ chessboard. There are three cases to examine:
1) the nth column is covered by two $1\times 1$ square tiles --- this case provides $a_{n-1}$ tilings:
2) the $(n-1)st$ and nth columns are tiled with one $1\times 1$ square tile and one larger tile --- this case accounts for $4a_{n-2}$ tilings; and
3) the $(n-2)nd$, $(n-1)st$, and $n$th columns are tiled with two of the larger tiles -- this results in $2a_{n-3}$ tilings.
These three cases cover all possibilities and no two of the cases have anything in common, so $a_n=a_{n-1}+4a_{n-2} + 2a_{n-3}$, $n\geq 4$, $a_1=1$, $a_2=5$, $a_3=11$.
Questions:
1) I have been going through a few examples and they have always started to count $a_1$, $a_2$ and possibly $a_3$. I have no idea why, but in some cases they are happy with just $a_1$ and $a_2$ and make some argue about these observations, conclude all possible cases have been covered and then come up with a recursive formula. But in other cases they, as in this particular case, consider 3 different cases before they conclude a formula. So how do I know when enough is enough? How do I know if I should make 2 observations, 3 observations or why not 4 observations? I have a really hard time solving problems because of this.
2) I have no idea what happens when they consider $n\geq 4$. Why FOUR times $a_{n-2}$, why TWO times $a_{n-3}$. To be honest I am not sure I understand any of the points 1), 2) or 3).
Here is a problem I tried to solve
Problem: Suppose the poker chips come in four colors - red, white, green, and blue. Find and solve a recurrence relation for the number of ways to stack n of these poker chips so that there are no consecutive blue chips.
I started to count for $n=1,2$ and $3$. But then I asked myself "is this enough to conclude a formula? If so, what now? What am I supposed to do?" I tried to reason in a similar manner like they did in the example but unfortunately I never managed to come up with a formula.
Hope someone can help me out. Perhaps some tips or tricks how I should think? Would also be happy if you could explain my questions for the example. I can't go on and keep reading the book before I got my head around this topics. Because there are many ideas which builds on recurrence relation. I have been thinking on this for 3 days now... sadly. Thanks for your time :)