Modular Reduction of polynomials in GF(2^m)
I don't know if this can help you, but did you know if in Binary Arithmetic, you have: $f(z)=z^m+r(z)$; then $z^m =f(z)+r(z)$.
NB: The addition here represents the bitwise XOR, that you can easily represent in hardware by appropriate gates.
Then Let's continue by examining the reduction algorithm you show us:
$c(z)=c_{2.m-2}.z^{2.m-2}+c_{2.m-1}.z^{2.m-1}+\cdots+c_{m}.z^{m}+c_{m-1}.z^{m-1}+\cdots+c_1.z+c_0$
We factorize by $z^m$; then, we obtain:
$c(z)=(c_{2.m-2}.z^{m-2}+c_{2.m-1}.z^{m-1}+\cdots+c_{m}).z^{m}+c_{m-1}.z^{m-1}+\cdots+c_1.z+c_0$
And replacing $z^m=f(z)+r(z)$ and performing the reduction mod f(z): the remaining is:
$c(z)=(c_{2.m-2}.z^{m-2}+c_{2.m-1}.z^{m-1}+\cdots+c_{m}).r(z)+c_{m-1}.z^{m-1}+\cdots+c_1.z+c_0$
Which can be computed, with the precomputed table!

Another interpretation of $r(z)$ from $z^m =1.f(z)+r(z)$ shows that $r(z)$ is the remainder of the division of $z^m$ by  $f(z)$ while the quotient is the constant polynomial 1.