This is essentially an issue of first quantization vs. second quantization. In second quantization we define the action of the time-reversal operator on the fermionic creation and annihilation operators to be
\begin{equation}
\begin{split}
\hat{\mathcal{T}}\hat{c}_{i, \uparrow}\hat{\mathcal{T}}^{-1} = \hat{c}_{i, \downarrow}, \quad \hat{\mathcal{T}}\hat{c}^{\dagger}_{i, \uparrow}\hat{\mathcal{T}}^{-1} = \hat{c}^{\dagger}_{i, \downarrow} \\
\hat{\mathcal{T}}\hat{c}_{i, \downarrow}\hat{\mathcal{T}}^{-1} = -\hat{c}_{i, \uparrow}, \quad \hat{\mathcal{T}}\hat{c}^{\dagger}_{i, \downarrow}\hat{\mathcal{T}}^{-1} = -\hat{c}^{\dagger}_{i, \uparrow}.
\end{split}
\end{equation}
where $\hat{c}^{\dagger}_{i, \sigma}/\hat{c}_{i, \sigma}$ are the fermionic creation and annihilation operators acting on site $i$ and spin state $\sigma=\uparrow/\downarrow$.
In addition $\hat{\mathcal{T}}$ is an antiunitary operator, therefore $\hat{\mathcal{T}}i\hat{\mathcal{T}}^{-1} = -i$ (we can show this by considering the Heisenberg uncertainty relationship and using the relations $\hat{\mathcal{T}}\hat{x}\hat{\mathcal{T}}^{-1} = \hat{x}$ and $\hat{\mathcal{T}}\hat{p}\hat{\mathcal{T}}^{-1} = -\hat{p}$).
We can summarise the actions of $\hat{\mathcal{T}}$ on the Fock space by first converting $\hat{c}_{i, \uparrow}, \hat{c}_{i, \downarrow}, \hat{c}^{\dagger}_{i, \uparrow}, \hat{c}^{\dagger}_{i, \downarrow}$, ... to $\hat{\psi}_{1}, \hat{\psi}_{2}, \hat{\psi}_{3}, \hat{\psi}_{4}$...
The action of $\hat{\mathcal{T}}$ is then surmised as $\hat{\mathcal{T}}\hat{\psi}_{A}\hat{\mathcal{T}}^{-1} = \sum_{B} U_{A, B}\hat{\psi}_{B}$ where $A, B$ are indices labelling the site and other relevant quantum numbers such as spin and $U$ is some unitary matrix. See Kitaev's paper for more details https://arxiv.org/abs/0901.2686. It should be emphasised that, so far, we have defined the action of time-reversal on the Fock space in the formalism of second quantization.
To move to the first quantized picture, we write the second quantized Hamiltonian, $\hat{H}$ in terms of the single-particle Hamiltonian as
\begin{equation}
\hat{H} = \sum_{A, B} \hat{\psi}_{A}^{\dagger}H_{A, B}\hat{\psi}_{B}
\end{equation}
where the operators $\hat{\psi}^{\dagger}_{A}/\hat{\psi}_{A}$ satisfy the usual anticommutation relations and $H$ is the first quantized Hamiltonian (basically just an $N\times N$ matrix). If the second quantized Hamiltonian possesses time reversal symmetry then $\hat{\mathcal{T}}\hat{H}\hat{\mathcal{T}}^{-1} = \hat{H}$. Using $\hat{\mathcal{T}}\hat{\psi}_{A}\hat{\mathcal{T}}^{-1} = \sum_{B} U_{A, B}\hat{\psi}_{B}$ we find
\begin{equation}
\begin{split}
\hat{\mathcal{T}}\hat{H}\hat{\mathcal{T}}^{-1} &= \sum_{A, B} \hat{\mathcal{T}} \hat{\psi}^{\dagger}_{A} \hat{\mathcal{T}}^{-1} \hat{\mathcal{T}} H_{A, B} \hat{\mathcal{T}}^{-1} \hat{\mathcal{T}} \hat{\psi}_{B} \hat{\mathcal{T}}^{-1} \\
&= \sum_{A, B}\sum_{C, D}U^{*}_{A, C} \hat{\psi}^{\dagger}_{C} H^{*}_{A, B} U_{B, D} \hat{\psi}_{D} \\
&= \sum_{C, D} \hat{\psi}^{\dagger}_{C} H_{C, D} \hat{\psi}_{D} = \hat{H}
\end{split}
\end{equation}
where $H_{C, D} = U^{*}_{A, C} H^{*}_{A, B} U_{B, D}$, i.e. $H = U^{\dagger}H^{*}U$. Here, the single particle Hamiltonian is complex conjugated since $\hat{\mathcal{T}}$ acts on the numerical parameters and reverses the sign of i. Therefore, we may define a first quantized version of $\hat{\mathcal{T}}$ which acts on the single particle space
\begin{equation}
T = \hat{\mathcal{T}}_{first quantized}
\end{equation}
We may then rewrite the action of time reversal on the first quantized Hamiltonian as
\begin{equation}
THT^{-1} = H \quad \text{where} \quad T = UK
\end{equation}
In summary, complex conjugation acts on numbers rather than operators.
For more information on the time reversal operator and its use in classifying topological phases of matter see these excellent review papers:
https://arxiv.org/abs/0912.2157
https://arxiv.org/abs/1512.08882
https://doi.org/10.1103/RevModPhys.88.035005
As you point out, there's an issue with the notion of quantum integrability defined, in analogy with the classical case, through the existence of sufficiently many conserved charges. For more on this, see Caux and Mossel, "Remarks on the notion of quantum integrability" https://arxiv.org/abs/1012.3587
So instead one can ask for a underlying algebraic notion, called 'Yang--Baxter integrability', that is well defined and implies the existence of many conserved charges. For more about this, see also https://physics.stackexchange.com/a/780318/.
The commuting charges constructed in this way (as logarithmic derivatives of the transfer matrix) are less and less local, see How local are the conserved charges in a quantum integrable model?. However, one can construct so-called quasi-local charges as the (first) logarithmic derivative of transfer matrices with higher spin in the auxiliary space (rather than higher derivatives of the 'fundamental' transfer matrix with spin-1/2 auxiliary space). See Ilievski, Medenjak and Prosen, "Quasilocal conserved operators in isotropic Heisenberg spin 1/2 chain" (https://arxiv.org/abs/1506.05049) and
Ilievski et al, "Complete Generalized Gibbs Ensemble in an interacting Theory" (https://arxiv.org/abs/1507.02993).
Best Answer
By $H^{(m)}$ the authors denote the usual conserved charge, i.e. the $m$th logarithmic derivative of the transfer matrix at a suitable value of the spectral parameter. For example, for $m=1$ this gives the XXZ spin chain Hamiltonian: a sum of terms $h^{(1)}_j$ that each act at blocks of at most one site adjacent to site $j$, i.e. at nearest neighbours $j,j+1$. This is true for any system size $N$.
If you similarly compute the next ultralocal charge $H^{(2)}$ you will get a sum of terms that act at sites $j,j+1,j+2$. (You might want do this explicitly as an exercise. The result can be found e.g. in my lecture notes on the arXiv.) Again this form is the same for any $N$.
In general, for fixed $m$ the summands act nontrivially at at most $m+1$ adjacent sites, independently of $N$. There's nothing extensive about these summands when you fix $m$.
The only thing that changes with $N$ is how many summands there are (the sum runs up to $N$), and how many such charges you have. Indeed, up to a possible common factor (depending on the normalisation of the R-matrix) the entries of the transfer matrix are polynomials in the spectral parameter of degree at most $N$, so you can in principle compute nontrivial charges for $0\leq m\leq N$. (Here $m=0$ corresponds to the momentum operator, i.e. the log of the translation = cyclic shift operator.)