Solution:

Extend sides $A B$, $C D$, $E F$, $G H$ to form a rectangle: let $X$ be the intersection of lines $G H$ and $A B$; $Y$ that of $A B$ and $C D$; $Z$ that of $C D$ and $E F$; and $W$ that of $E F$ and $G H$.

As $B C=2$, we have $B Y=Y C=\sqrt{2}$. As $D E=4$, we have $D Z=Z E=2\sqrt{2}$. As $F G=2$, we have $F W=W G=\sqrt{2}$.

We can compute the dimensions of the rectangle: $W X=Y Z=Y C+C D+D Z=3+3\sqrt{2}$, and $X Y=Z W=Z E+E F+F W=2+3\sqrt{2}$. Thus, $H X=X A=X Y-A B-B Y=1+2\sqrt{2}$, and so $A H=\sqrt{2} H X=4+\sqrt{2}$, and $G H=W X-W G-H X=2$. The perimeter of the octagon can now be computed by adding up all its sides.