I do not understand here on the page $8$
that in the definition of traceless tensor product there are 3 indices $i,j,k$
in the rightmost term but in the preceding term and in front of the equality sign $=$
both have just two indices $i,j$.
Also, is there used the Einstein summation convention in this formula ?
Lastly, what is the fraction in front of the rightmost term intended to mean $-\frac{1}{2}\delta_{ij}\cdot$
Suppose the dimension is $N$ (in the paper, $N=2$, but we work more generally).
$\delta_{ij}$ is the Kronecker delta. Its components are given by $\delta_{ij}=1$ if $i=j$ and $0$ if $i\neq j$. So according to page $8$ of the linked paper, $f\hat{\otimes}g$ has $i,j$ component $ f_ig_j-\frac{1}{N}\delta_{ij}f_kg_k$, using the summation convention. This is equal to $$(f\hat{\otimes}g)_{ij} = \left\{\begin{array}{cc} f_ig_i - \frac{1}{N}\sum_{k=1}^N f_kg_k & : i = j \\ f_ig_j & : i \neq j. \end{array}\right.$$ That's because the Kronecker delta is only non-zero when $i=j$.
To get the trace, we sum \begin{align*} \sum_{i=1}^N (f\hat{\otimes} g)_{ii} & = \sum_{i=1}^N\Bigl[f_ig_i - \frac{1}{N}\sum_{k=1}^N f_kg_k\Bigr] \\ & = \sum_{i=1}^N f_ig_i - \sum_{k=1}^N \frac{1}{N}\sum_{i=1}^N f_kg_k \\ & = \sum_{i=1}^N f_ig_i - \sum_{i=1}^N f_ig_i=0.\end{align*}
Imagine subtracting $1/N$ of the trace down the diagonal, so that when you sum down the diagonal, each of the $\text{trace}/N$ pieces sums up to subtracting the trace once.