tensor product in a k-linear category Studying this paper i met a "tensor product" construction.At lemma 2.4.1 it is defined an object in D as a direct sum of  $$E[n]\otimes_k H(E[n])$$ I searched on the internet and i found a comment here that says it is a finite direct sum of $dimH(E[n])$ many copies of $E[n]$.But even with this assumption i have no idea of how to define the canonical natural transformation $\zeta_{1}$ and i was wondering if there is a precise definition of this kind of tensor product which maybe i can deal better with(and of course to define $\zeta_{1}$).I have a basic knowledge on homological algebra and i've never met this notion on  $k$-linear triangulated categories.

Remark: $h_{A_{1}}$ in the paper is a notation for the representable functor $Hom(-,A_{1})$

 A: Rough background on enriched categories: $\newcommand\V{\mathcal{V}}\newcommand\C{\mathcal{C}}$
Very roughly a $\V$-enriched category is a category $\C$ whose hom-sets have extra structure so that they are objects belonging to another category $\V$, which has a monoidal product bifunctor,
often denoted $\otimes : \V\times \V\to \V$ such that composition of maps can be thought of 
as giving an arrow in $\V$, $$\circ_{X,Y,Z}: \C(Y,Z)\otimes \C(X,Y)\to \C(X,Z).$$
In your case $\V=k\textrm{-}\mathbf{Vect}$, with the usual tensor product,
since all the hom-sets are $k$-vector spaces and composition is bilinear, so the 
composition induces a map $\C(Y,Z)\otimes \C(X,Y)\to \C(X,Z)$, as required.
(Also we'll need $\V$ to be closed monoidal for our purposes, which in the case of vector spaces is just the fact that $\V(V_1,V_2)$ is again a vector space.)
Tensored categories
We say that a $\V$-enriched category $\C$ is tensored over $\V$ if there is a functor 
(usually also denoted $\otimes$, but for which I'll use $\boxtimes$ for clarity here)
$$\boxtimes : \V\times \C\to \C $$
such that we have a natural isomorphism
$$ \C(v\boxtimes c_1,c_2) \cong \V(v,\C(c_1,c_2)).
$$
In other words, it satisfies a sort of tensor-hom adjunction, and makes $\C$ into a module category over $\V$. See the nlab for more on tensored categories, although the nlab calls them copowered categories. Now you can see that this question is also related
The explanation
As you may have guessed, what's going on here is that we are taking a tensor product in the sense just defined. These are not always guaranteed to exist, but in the special case of vector spaces, we just need coproducts to exist, as the comment you linked to indicates. We also don't necessarily get all tensor products, so we don't necessarily have a functor as we would like, the construction I'll give only goes through for finite vector spaces in your example.
First let's say what it means for a tensor to exist without the whole functor existing.
We say that an object $y\in \C$ is the tensor of $v\in \V$ with $c_1\in\C$ if
there is a natural isomorphism (in $c_2$)
$$ \C(y,c_2) \cong \V(v,\C(c_1,c_2)).$$
Note that the natural isomorphism is actually part of the definition of $y$. $y$ is then unique up to unique isomorphism if it exists, and I will denote such a $y$ by $v\boxtimes c_1$.
If such objects $y$ exist for all pairs $v$ and $c_1$, then they assemble into the functor $\boxtimes : \V\times \C\to \C$ discussed above.
Now let $\V$ be the category of $k$-vector spaces, and $\C$ be a $k$-linear category (meaning enriched over $\V$) with finite coproducts and a zero object (comes for free with triangulated category, since those are additive).
Then if $v\in\V$ is finite dimensional, $v\boxtimes c$ exists for all objects $c\in\C$.
Proof
Take a basis $e_1,\ldots, e_n$ for $v$. Let $y = \bigoplus_{i=1}^n c$, then for any $d\in\C$, 
$$\C(y,d) 
= \C\left(\bigoplus_{i=1}^n c,d\right)
\cong 
\bigoplus_{i=1}^n \C(c,d)
\cong 
\V(v,\C(c,d)).\quad \blacksquare
$$
What is $\zeta_1$?
We have 
$$A_1 := \bigoplus_n E[n]\boxtimes H(E[n]),$$
where $H:\C^{\text{op}}
\to\V$ is a contravariant cohomological functor of finite type 
(meaning that $\bigoplus_n H(E[n])$ is finite dimensional for all $E\in\C$).
Fix bases $v_{n,i}$ for $1\le i \le m_n=\dim H(E[n])$, so that 
$$A_1 = \bigoplus_n \bigoplus_{i=1}^{m_n} E[n].$$
We want to show that there is a canonical natural transformation (natural in $F\in\C$)
$$
\C(F,A_1) \to H(F).
$$
Now notice that 
$$
\C(F,A_1)
\cong
\bigoplus_n\bigoplus_{i=1}^{m_n}\C(F,E[n])
\cong 
\bigoplus_n \C(F,E[n])\otimes_k H(E[n])
$$
since the sum is finite. Note that the tensor product is now the usual 
tensor product over $k$.
Then for each $n$, we have a bilinear map $\C(F,E[n])\otimes_k H(E[n])\to H(F)$
defined by $\phi\otimes h \mapsto H(\phi)h$. Together these give the desired map $\zeta_1$.
Observe that $\zeta_1$ is surjective for $F=E[n]$, since we can just observe that 
the image of $\mathrm{id}_{E[n]}\otimes h$ is $h$ for all $h\in H(E[n])$.
