A $3$-uniform hypergraph $H=(V,E)$ is a hypergraph where each hyperedge is a $3$-element subset of $V$. So, let $H$ be a $3$-uniform hypergraph on $n$ vertices.
Problem: I want to show that if every six vertices of $H$ contain at most $2$ hyperedges then $H$ has $\omicron(n^{2})$ hyperedges.
My main idea is to consider a simple graph $G$ with vertex set the same as $H$, i.e. $V(G):=V$, and if three vertices $u,v,w\in V$ are connected with an edge in $H$, i.e. $\{u,v,w\}\in E$, I connect them in $G$ via a triangle, i.e. $\{u,v\},\{u,w\},\{v,w\}\in E(G)$. Hence, I have reduced the problem into showing that the number of triangles in $G$ is $\omicron(n^{2})$, which to me seems more accessible.
Attempt 1: Let $t$ be the number of triangles in $G$ and suppose that $t\geq \epsilon n^{2}$ for some $\epsilon>0$. First, note that by assumption every edge of $G$ is contained in at most $2$ triangles. Then, $G$ is $\epsilon/2$-far from being triangle free, i.e. if we remove fewer than $\epsilon n^{2}$ edges, the graph $G$ still contains a triangle. Then, by the Triangle Removal Lemma, $G$ contains at least $\delta n^{3}$ triangles, but I can show that $t$ is $\omicron(n^{2})$, as every edge is in at most $2$ triangles. I feel that this proof is a little weird and so I am a bit worried of this
Attempt 2: I think that every block of $G$ is isomorphic to $K_{4}^{-}$, i.e. a square with a diagonal, or to $K_{3}$, but I am finding it hard to prove it, since the construction of the graph $G$ is a little complicated to me. If this holds then $G$ has at most $2(n/4)$ triangles as there at at most $n/4$ such blocks and every such block contains at most $2$ triangles.
Although it has been three years, I hope this answer can help someone else. There is a proof here, but I have a different proof as follows.
Take another graph $G=(V,F)$ on the same vertex set. We draw an edge between $x,y$ if there is some hyperedge $e\in E$ such that $\{x,y\}\subset e$. Now we consider what does a triangle in $G$ mean in $H$. Suppose $(x,y,z)$ forms a triangle in $G$, then there are two possibilities, that $\{x,y,z\}\in E$ or not. In the first case, the three edges between $x,y$, and $z$ are trivially existing in $F$, while in the second case it implies there is some other(possibly distinct) points $w_1,w_2,$ and $w_3$ such that $\{x,y,w_1\},\{x,z,w_2\},\{y,z,w_3\}\in E$, implying there are three hyperedges in the set $|\{x,y,z,w_1,w_2,w_3\}|\le 6$. Therefore, for a hypergraph $H$ with the assumed property, its corresponding graph $G$ has only triangles of the first kind, implying the number of triangles in $G$ equals $|E|$. By triangle removal lemma, it remains to show $|E|=o(n^3)$.
Now note that if there are two hyperedges $e_1,e_2$ such that $|e_1\cap e_2|=2$, then these four vertices in $e_1\cup e_2$ cannot be in any other hyperedge due to the constraint. In other words, we can decompose $V$ into two disjoint parts $V_1,V_2$ such that $V_1$ is the union of those four vertices systems and every two induced hyperedges in $V_2$ has at most one vertex in common. Moreover, there is no hyperedge that contains vertices in both $V_1$ and $V_2$. Therefore, we can count the number of hyperedges in these two sub-hypergraphs separately.
In the first case, since any such four vertices system is disjoint from the other, the total number of hyperedges is at most $2(|V_1|/4)\le n$. In the second case, for any vertex $v$ contained in some hyperedges, the rest two-points-systems are disjoint by the definition of $V_2$. This implies there are at most $(|V_2|-1)/2\le n/2$ hyperedges containing $v$. Note each hyperedge contains $3$ vertices, we have the number of hyperedges in $V_2$ is at most $n(n/2)/3=O(n^2)$. In summary, we have $|E|=O(n^2)$, completing the proof.