64 bits password = 13 characters? If we use upper and lower case letters, and 10 digits, we get
  approximately 6 bits per character. Then, strings of 13 characters
  should work.


I saw above explanation in the material, but I cannot understand how 13 characters are appeared. How can I compute?
Suppose the number of a set $S$ is denoted as $Card(S)$ (which is sort for cardinality if you didn't know), then 

$Card($uppercase-letters$)+Card($lowercase-letters$)+Card($digits$)
=26+26+10=62$

To represent an element from a set of 62, you need a string of at least 6 bits, this is because $2^{6} = 64 \ge 62$

However I don't understand the title where it says "64 bits password = 13 characters" because $ 6 \cdot 13 = 78 $ which is way bigger here, than 64. 

To achieve at least 64-bit entropy, you need to uniformly draw ${ {64} \over {log_2{62}} } \approx 11.00$ characters (10.749 to be more exact).