Why is it an equivalent definition of a triangulated full subcategory? We know that an additive full subcategory S of a triangulated category T is called a triangulated subcategory if it is closed under isomorphism, shift and if any two objects in a distinguished triangle in T are in S, then so is the third. See Neeman's book section 1.5. Why is it equivalent to that the inclusion functor is exact in the sense that it preserves distinguished triangles? And what if assuming any additive subcategory which may not be full?
 A: A triangulated functor $F\colon \mathcal T'\to \mathcal T$ must preserve structures. So we have $F$ commutes with translations (aka shifts/suspensions) and takes triangles in $\mathcal T'$ to triangles in $\mathcal T$. 
If $\mathcal T$ is a triangulated category and $\mathcal T'$ is a full additive subcategory of $\mathcal T'$, we ask when the inclusion functor $i\colon \mathcal T'\hookrightarrow \mathcal T$ gives us an induced triangulated subcategory as its essential image. By induced, we mean a definition of triangles and translation that comes from restrictions of translation and triangles in $\mathcal T$. 
So let's look at why these three criteria are reasonable:
(1) Closed under isomorphisms: Since the substructure is given by the essential image, we should have $\mathcal T'$ closed under isomorphism. 
(2) Closed under translations: Clearly, if the translation of $\mathcal T'$ is inherited from $\mathcal T$ the restriction to $\mathcal T'$ must factor through itself.
(3) 2 of 3: If $\mathcal T'$ has triangles consisting of all those triangles in $\mathcal T$ such that all objects of the triangle are in $\mathcal T'$, we need only consider two of three since cones of morphisms in $\mathcal T'$ must be in $\mathcal T'$ (it is triangulated) and translation allows us to consider any two. 
