Overall, I've counted
so there will inevitably be some degree of case bashing.
This is how I would proceed.
Notice firstly that
None of the digits is zero since that would make some of the others equal or negative.
Suppose $f < 9$
Then either $d$ or $e$ is less than $4$ (since they are distinct). Since both are the sum of distinct digits, it must be that the smaller of the two is $3$ and the other is $4$ or $5$.
Firstly, suppose that this $d=3$. Then $a$ and $b$ must be $1$ and $2$ in some order. Since $e$ must be $4$ or $5$ its clear that $b$ cannot be $2$ (since $c$ would have to be $2$ or $3$) and if $b=1$ then $c=4$ and $e=5$ is the only possible solution. This gives us the first solution $$(a,b,c,d,e,f) = (2,1,4,3,5,8)$$
Since the equations remain the same if we swap $d$ with $e$ and $a$ with $c$, we quickly obtain a second solution which comes from assuming $e=3$ instead $$(a,b,c,d,e,f) = (4,1,2,5,3,8)$$ and these are all the solutions in this branch
Now suppose $f=9$
Then $d$ and $e$ are either $3$ and $6$ or $4$ and $5$, in some order (since we've established that both $d$ and $e$ must be at least $3$).
$d=3$ gives us that $a$ and $b$ are $1$ and $2$ in some order.
With $b=2$ we must have $c=4$ and this gives another solution $$(a,b,c,d,e,f) = (1,2,4,3,6,9)$$
With $b=1$ we must have $c=5$ and we have yet another solution $$(a,b,c,d,e,f) = (2,1,5,3,6,9)$$
Again, we can reverse the role of $(d,e)$ and $(a,c)$ to retrieve the solutions in the $e=3$ branch $$(a,b,c,d,e,f) = (4,2,1,6,3,9)$$ $$(a,b,c,d,e,f) = (5,1,2,6,3,9)$$
Finally, $d=4$ gives us $a$ and $b$ being $1$ and $3$ in some order.
$b=1$ doesn't work because it would mean $c=4=d$.
$b=3$ gives us another solution $$(a,b,c,d,e,f) = (1,3,2,4,5,9)$$ and swapping $(d,e)$ and $(a,c)$ gives us the last solution which comes from the $e=4$ branch $$(a,b,c,d,e,f) = (2,3,1,5,4,9)$$