I dont understand how i can get the following expression: $$\varepsilon_{ijk}\varepsilon_{ijk'}=2\delta_{k k'}$$ I try the following and end up with a $-2$ instead of a $2$:
$$\varepsilon_{ijk}\varepsilon_{ijk'}=\delta_{ij}\delta_{jk'}-\delta_{ik'}\delta_{jj}=\delta_{ik'}-3\delta_{ik'}=-2\delta_{ik'}$$
You are probably trying to use the well known identity $$\varepsilon_{ijk}\varepsilon_{hlk}=\delta_{ih}\delta_{jl}-\delta_{il}\delta_{hj}\tag{1}$$ Notice that the index we sum over (i.e. $k$) is the last index in $(1)$.
In your expression we sum over the first two indices. In order to use the formula we need to rearrange the indices using the skew symmetric property of Levi-Civita symbol. For convenience we "move" the index $k$ two steps to the left:
$$\varepsilon_{ijk}=-\varepsilon_{ikj}=\varepsilon_{kij}\tag{2}$$
Of course we can do the same with $k^\prime$ $$\varepsilon_{ijk^\prime}=-\varepsilon_{ik^\prime j}=\varepsilon_{k^\prime ij}\tag{3}$$ Now using $(2),(3)$ and applying $(1)$ with contraction over the last index $j$ we have $$\varepsilon_{ijk}\varepsilon_{ijk^\prime}=\varepsilon_{kij}\varepsilon_{k^\prime ij}=\delta_{kk^\prime}\delta_{ii}-\delta_{ki}\delta_{ik^\prime}$$ $$=3\delta_{kk^\prime}-\delta_{kk^\prime}=2\delta_{kk^\prime}$$