I want to know if there are any systematic properties of $(a,b,c,d)$ that guarantee the inequality
$$2\max\{a+b,0\}+2\max\{c+d,0\}>\max\{a+b+c+d,0\}+\max\{a,0\}+\max\{b,0\}+\max\{c,0\}+\max\{d,0\}.$$
Or equivalently,
\begin{align}&\max\{a+b,0\}+\max\{c+d,0\}-\max\{a+b+c+d,0\}\\>&\max\{a,0\}+\max\{b,0\}+\max\{c,0\}+\max\{d,0\}-\max\{a+b,0\}-\max\{c+d,0\}.\end{align}
We have $a,b,c,d\in (-1,1)$. The inequality does not necessarily hold. For example, if $a=-b$ and $c=-d$, then the opposite inequality should hold.
However, if $a,b>0$ and $c,d<0$, we have the inequality. Can we find a general condition under which the inequality holds?
In general, you can think of the 7 max-functions as partitions of the 4d-space into halfspaces. E.g. $\max\{a,0\}$ decides that the inequality in question looks different in the halfspace $a>0$ than in the halfspace $a \le 0$. So you would need to check all intersections of these halfspaces, this is $2^7 = 128$ different conditions (partitions of the space) in each of which the inequality in question looks different.
To reduce this number of conditions, exploit symmetries and implications.
For symmetries, you can exploit that $a$ and $b$ can be exchanged and the inequality in question looks the same. That also holds for exchange of $c$ and $d$. Further, it holds for the interchange of $(a,b)$ with $(c,d)$.
For implications, you can exploit that $a>0$ and $b > 0$ imply $a+b >0$, so this third inequality does not lead to another partition. Likewise for other inequalities.
Further, you can exploit that if all variables are less than (greater than) zero, then we have equality, and the inequality is disobeyed. So at least one variable has to be positive and you can choose that variable to be $d$. Likewise, at least one variable has to be negative. Due to homogeneity, you can actually choose $a,b,c,d\in (-1,1)$ which you already stated.
So you have to inspect whether $$ 2\max\{a+b,0\}+2\max\{c+d,0\}>\max\{a+b+c+d,0\}+\max\{a,0\}+\max\{b,0\}+\max\{c,0\}+d. $$
The $2^6 = 64$ remaining conditions reduce significantly under the above rules.
Again, for $a,b,c>0$ we have equality so the inequality is not true. Then by implication, none of the other inequalities need to be checked.
1.1. For $c > -d$ 1. reduces to $$ a+b+c+d + c >\max\{a+b+c+d,0\} $$ or $$ c >-\min\{a+b+c+d,0\} $$ and since $a+b>0$, $c+d>0$, this is $c >0$ which cannot hold since $c<0$ is inherited from 1.
1.2 For $c < -d$ this reduces to $$ a+b >\max\{a+b+c+d,0\} +d $$
1.2.1. For $a+b+c+d <0$ we have $a+b >d$ for the inequality to hold.
Just to be complete: the condition $c <-d$ is stronger than $c <0$, so some conditions become obsolete.
The full set of conditions is now:
$a,b,d>0$, $c <-d$,$a+b+c+d <0$, $a+b >d$.
Example: $(a,b,c,d) = (0.1,0.1,-0.5,0.1)$
1.2.2. For $a+b+c+d >0$ we have $c <-2 d$ for the inequality to hold.
The full set of conditions is now:
$a,b,d>0$, $c <-2d$, $a+b+c+d > 0$.
Example: $(a,b,c,d) = (0.3,0.3,-0.3,0.1)$
This completes case 1. I leave it to check
2. $a<0,b>0$ and $c <0$
3. $a,b <0$ and $c <0$
4. $a,b <0$ and $c >0$
and their subcases. There are only these 4 main cases, due to the above symmetries.