We can use a "big M" approach, assuming that you can find a positive constant $M$ such that $\vert y_{jk}-y_{ik} \vert \le M$ for all $k\in K$ and $(i,j)\in V.$
Introduce new binary variables $u_{ik}$ and continuous variables $v_{ijk}.$ The second objective summand (the mess in brackets) becomes $v_{ijk}.$ Now add the following constraints:
$$3u_{ik} + z_{ik} \le 3\, (1)$$
$$u_{ik} + z_{ik} \ge 1 \, (2)$$
$$v_{ijk} \le M u_{ik}\, (3)$$
$$v_{ijk} \ge -M u_{ik}\, (3')$$
$$v_{ijk} \le M x_{ijk}\, (4)$$
$$v_{ijk} \ge -M x_{ijk}\, (4')$$
$$v_{ijk} \le y_{jk} - y_{ik} + M(2 - u_{ik} - x_{ijk}) \, (5)$$
and
$$v_{ijk} \ge y_{jk} - y_{ik} - M(2 - u_{ik} - x_{ijk}) \, (5')$$
Constraints (1) and (2) combine to set $u_{ik} =1$ if and only if $z_{ik} = 0.$ Constraints (3) and (3') say that the $(i,j,k)$ objective term is 0 if $u_{ik}=0$ (meaning $z_{ik} \in \lbrace 1,2,3 \rbrace$). Constraints (4) and (4') say that the objective term is 0 if $x_{ijk} =0.$ Finally, constraints (5) and (5') say that if $u_{ik}=1$ (meaning $z_{ik}=0$) and $x_{ijk}=1,$ then the objective term equals $y_{jk}-y_{ik}.$
Addendum: If the solver supports "if-then" (implication) constraints (CPLEX does), then the "big M" constraints can be replaced with implications, as follows.
$$u_{ik}=0 \implies v_{ijk}=0$$
$$x_{ijk}=0 \implies v_{ijk}=0$$
$$u_{ik} + x_{ijk} = 2 \implies v_{ijk} = y_{jk}-y_{ik}$$
As a general rule, you are probably better off with the "big M" constraints if you can find a reasonably tight value of $M.$