There are $5$ combinations of $4$ numbers from $1-9$. If you write these on two coins, so there are $2$ numbers on each coin. Then you flip the coin. The sum of the two numbers add to $8,9,10,$ or $11.$
In other words, you can combine $2$ numbers to make $8,9,10,$ or $11.$
What are the combinations
Let one coin have numbers $a$ and $b$ with $a<b$; and the other coin, $c$ and $d$ with $c<d$.
We then need $a+c=8$ and $b+d=11$. Also $a+d$ is $9$ or $10$, and $b+c$ is respectively $10$ or $9$. Without loss of generality, assume that $a+d=9$, and so $b+c=10$.
Solving the system $\left\{\begin{array}{c}a+c=8\\ a+d=9\\ b+c=10\\b+d=11 \end{array}\right.$
with the constraints that $a,b,c,d$ are different integers from $1$ to $9$, should give you all the possibilities. (Note that once a value of say $a$ is chosen, the other values are all determined.)