Be careful of the distinction between rolling below a number, and rolling that number or lower. As @calculus noted, the probability of rolling $13$ or lower is $13/20 = 65$ percent, not $60$ percent.
One will roll $n$ or higher at normal odds with probability $(21-n)/20$. One rolls $n$ or higher at advantage with probability $1-(n-1)^2/400$. And one rolls $n$ or higher at disadvantage with probability $(21-n)^2/400$. For $n = 13$, this yields $8/20 = 40$ percent, $1-144/400 = 256/400 = 64$ percent, and $64/400 = 16$ percent, respectively. I'm not sure where the linked table got $63.9$ percent.
For your problem, your phases are equivalent to failing to roll a $14$ at normal (first two phases) and then failing to roll a $14$ at disadvantage (second two phases). Those probabilities are $1-(21-14)/20 = 1-7/20 = 13/20 = 65$ percent, and $1-(21-14)^2/400 = 1-49/400 = 351/400 = 87.75$ percent, respectively.
To compute the probability of joint events (e.g., one must pass phases $1$, $2$, and $3$ to reach phrase $4$), one multiplies the probabilities associated with each phase. That is, the probability of reaching phase $4$ is the product of the probabilities of passing phases $1$ through $3$; the probability of passing phase $4$ is the product of the probabilities of passing phases $1$ through $4$.
In other words, you would multiply $(13/20)(13/20)(351/400) \doteq 37.074$ percent probability of reaching your phase $4$ (i.e., failing the first three rolls), and $(13/20)(13/20)(351/400)(351/400) \doteq 32.533$ percent probability of passing through phase $4$ (i.e., failing all four rolls).