The problem is to construct a topological structure on the space of all triangles in the plane, such that each triangle has a neighborhood homeomorphic to $\mathbb{R}^6$.
I am confused by the space of all triangles and how to approach this problem. Is the space just all the 3-tuples such that three vertices are not co-linear?
Yes I think you have the right idea that a triangle in the plane is determined by three points which are not colinear. The $3$-tuple of vertices $(P_1, P_2, P_3)$ can be considered as an element of $(\mathbb{R}^2)^3\cong \mathbb{R}^6$, so consider the space $$ \Delta = \{ (P_1, P_2, P_3) \in \mathbb{R}^6\ |\ P_1, P_2, P_3\text{ are not colinear in }\mathbb{R}^2\} $$ with the subspace topology. Now you need to make sure that if $T\in \Delta$ is a triangle, then there is a small ball $B_\epsilon(T)\subset \mathbb{R}^6$ such that $T\in B_\epsilon(T) \subset \Delta$. If $h$ is the shortest altitude (i.e. the orthogonal distance from a vertex to the opposite edge) in $T$ then I think a ball of radius $\epsilon = \frac{h}{3}$ will suffice.
However there is some ambiguity in "what is a triangle?" Does re-ordering the vertices count as a different triangle, or should it be the same? If the order of the vertices matters then what I wrote above is an answer. However, if we want different vertex orderings to determine the same triangle then we need to think about the action of the permutation group $S_3$ on $\Delta$. In particular, given a $\sigma\in S_3$ we let it act by $\sigma\cdot(P_1, P_2, P_3)=(P_{\sigma(1)}, P_{\sigma(2)}, P_{\sigma(3)})$, and the space of unordered triangles can be defined by
$$\Delta' = \Delta/S_3 $$
We need to verify that this quotient space still locally looks like $\mathbb{R}^6$. Notice that this action is free on $\Delta$, by which I mean given any $T\in \Delta$ and $\sigma\neq 1 \in S_3$ we have $\sigma\cdot T \neq T$, because the points of $T$ cannot be equal to each other (since they are not colinear). Now the general theory of Covering Spaces kicks in and tells us that since $S_3$ is finite discrete and its action is free then the quotient map $q\colon \Delta \to \Delta'$ is a covering map, and covering maps preserve local neighbourhood structure. The details in this case would go roughly as follows: Note that since $S_3$ acts by isometries we have $\sigma\cdot B_\epsilon(T) = B_\epsilon(\sigma\cdot T)$, and by choosing $\epsilon$ small enough as above we can ensure that these $6$ different $\epsilon$ balls are all disjoint. If we let $T'=q(T)$ (i.e. it is the $S_3$ orbit of $T$) then the definition of the quotient topology tells use that $q(B_\epsilon(T))$ is an open neighbourhood of $T'$. I will leave it as an exercise to show that for each $\sigma$ the restriction $q|_{\sigma\cdot B_\epsilon(T)}$ is a homeomorphism.