Look at my relabeling of your diagram. A path (I) could be $ADB$, that is $A\to D\to B$ from A thru D to B.
Then path (II) is $AED$. On path (I) there is only one possibility $VVVVVHHHHHHHH$, go vertical five times to $D$ then go horizontal eight times to $B$. Doing that is a path $A\to B$. Note that means $C$ is on path (I) but $E$ is not.
Now consider path (II), $AEB$ or $A\to E\to B$. There at two parts $A\to E$ & $E\to B$.
The first part consists of $\bf{HVVV}$ in some order.
There are four ways to order that string.
$C$ may or may not be on that part, but $D$ is not.
The second part of path (II) consists of $\bf{VHHHHHHH}$ in some order There are eight ways to order that string.
$F\text{ or }G$ may or may not be on that part, but not both.
The total is $1+4\cdot 8$. Now you explain to others why!