I'm not sure how the notation you're using matches up to different presentations of this equivalence. I'm going to assume $E$ is nearby cycles, $F$ is vanishing cycles, $c$ is the obvious map from nearby cycles to vanishing cycles arising from the mapping cone, and $v$ is the less-obvious map. 

I'll write each operator as sending $(E, F, c,v)$ to $(E', F', c', v')$

The stalks at the special point and geometric generic point are both preserved by pullback, so the mapping cone $ E \to_c F$ of the natural map between them is preserved by pullback. Thus $E'=E, F'=F, c'=c$. However, $v$ might be changed. Indeed, the monodromy operators $cv+1$ and $vc+1$ are raised to their $k$th power by pullback along $f_k$. Since the $k$'th power is $\sum_{i=0}^k \binom{k}{i} (cv)^i$ or $\sum_{i=0}^k \binom{k}{i} (vc)^i$ respectively, to obtain these as $c'v'+1= cv'+1$ and $v'c'+1= v'c+1$, we must have $$v' = \sum_{i=1}^k \binom{k}{i} v(cv)^{i-1} =  \sum_{i=1}^k \binom{k}{i} (vc)^{i-1}v. $$

So $f_k^*$ sends $(E, F, c,v)$ to $(E, F, c, \sum_{i=1}^k \binom{k}{i} v(cv)^{i-1})$.

$f_k^!$ can be obtained by dualizing, applying $f_k^*$, and dualizing again. I guess dualizing sends $E$ to $E^\vee$, $F$ to $F^\vee$, $c$ to $v^T$ and $v$ to $c^T$. Applying this, we see $f_k^!$ sends $(E, F, c,v)$ to $(E, F,  \sum_{i=1}^k \binom{k}{i} c(vc)^{i-1},v)$.

Now pushforward along $f_k$ preserves the stalk at the special point but not the nearby cycles or vanishing cycles. Rather, it sends nearby cycles to a vector space of dimension $k$ times higher where $(1+v'c')$ acts by the $k$'th root. Thus, we set $E' = E^k$ where we want $1+v'c'$ to act by sending $(e_1,\dots, e_k)$ to $(e_2, e_3,\dots, e_k , (1+vc)e_1$. We can think of each copy of $E$ as representing the stalk at one of the $k$ points in the geometric generic fiber of $f_k$, and since the natural map from the stalk at the special fiber to the stalk at the geometric generic fiber is nontrivial at each of those points, we would like the kernel of $c'$ to include the kernel of $c$ written diagonally. This can be achieved by setting $F' = E^k \times F$ and letting $c'$ send $(e_1,\dots, e_k)$ to $(e_2-e_1,e_3-e_2, \dots, e_k-e_{k-1}, c e_1)$. This gives the correct kernel and cokernel.

To get the correct value for $1+v'c'$, we just need $v'$ to send $(f_1,\dots, f_k)$ to $(f_1, \dots, f_{k-1}, v f_k)$.

This gives, $f_{k*}$, and $f_{k!}$ is the same since $f_k$ is finite.

I think one can check this is correct using the adjunction, but I haven't worked it out.