There are two "cutoff" (for lack of a better word) scales here, $\Delta$ and $\alpha$. $\Delta$ is inherent to the fermionic theory, it is introduced to remove the divergence of the correlation function \begin{equation} \langle \psi^\dagger (x+\Delta) \psi(x)\rangle \end{equation} as $\Delta \to 0$. To get the (point-split) density $\rho(x) ={:}\psi^\dagger (x) \psi(x){:}$, we must subtract this divergence. The reason we have to do this is because we are interested in the density relative to the vacuum state, the filled Fermi sea denoted by $\vert 0 \rangle $. Alternatively, this can be done by normal-ordering the operator \begin{equation} \rho(x) \equiv \mathcal{N} \psi^\dagger (x) \psi(x) \end{equation} where normal-ordering is defined on the momentum-space creation and annihilation operators as \begin{equation} \mathcal{N} c^\dagger_k c_{k'} = - c_{k'} c^\dagger_k \quad \text{if } k = k' \in \text{ Fermi sea,} \end{equation} otherwise it acts trivially. One can check that, by this definition, $\langle 0 \vert \rho(x) \vert 0\rangle = 0$. One can check appendix G.2 of this paper for more on the connection between point splitting and normal ordering. 66
Therefore, using the point-splitting recipe, the density operator relative to the vacuum state is defined as \begin{equation} \rho(x) \equiv \lim_{\Delta \to 0} \big[ \psi^\dagger (x+\Delta) \psi(x) - \langle 0 \vert \psi^\dagger (x+\Delta) \psi(x) \vert 0 \rangle \big] \end{equation}
The other "cutoff" $\alpha$ describes the length scale to which our bosonization map is valid. The boson field $\phi(x)$ is made up of momentum space operators $b_q$ which in turn is made up of fermionic operators \begin{equation} b_q \sim \sum_k c^\dagger_{k+q} c_k . \end{equation} $\alpha^{-1}$ acts as a soft UV cutoff for the range of values of $q$ that we use when we write $\phi(x) \sim \sum_q e^{-iqx} b_q + h.c. $ So if $\alpha$ is our cutoff, then it does not make sense to use the bosonization map to talk about correlation functions of boson operators which are closer than $\alpha$ from each other. So, to get the density operator exactly, we must take $\alpha \to 0$ since it is a correlation function of operators at the same location.
Summary: $\alpha \to 0$ must be taken first as we want to calculate correlation functions of fermion operators at the same location, so we want all the UV information intact under the bosonization map. Once we have the exact representation of the fermion operators in terms of the bosons, $\Delta$ is used as a trick to calculate the fermion density relative to the Fermi sea, and is not a physical (or essential) parameter. One can do the same calculation without $\Delta$ using normal ordering.
EDIT: regarding OP's comment about thinking in terms of a lattice -- $\Delta$ is used as a trick to calculate the relative density operator in the continuum, therefore it does not make sense to take the limit $\Delta \to 0$ on the lattice (what does it mean to say $\psi(x+\Delta)$ when $\Delta$ is taken to be smaller than the lattice spacing?). This means that we should go to the continuum first by taking $\alpha \to 0$. 66