The answer to this depends on how the reagents are mixed. For one, you can have the reaction without water, as pointed out by @Klaus in the comments. If you blow dry HCl on dry NaOH, they will react without involving water in the reactants.
$$\ce{HCl + NaOH -> H2O + NaCl}$$
This is probably not the case though, and if you're dealing with solutions, then yes, water will react. If you start by adding the HCl to pure water, you'll have
$$\ce{HCl + H2O -> H3O+ + Cl-}$$
This representation is prefered over the simple dissociation $\ce{HCl -> H+ + Cl-}$ because it shows that you don't have "free" protons in solution. You could actually even say that this reaction has already happened since, considering you're using hydrochloric acid it's already dissolved, and this step would be just a dilution.
If you start with the NaOH solution, you could argue that water does not react since
$$\ce{NaOH -> Na+ + OH-}$$
but that only happens in the liquid state (usually in solution). Also, you can't ignore the ions solvation (which is one, if no the most important factor which affects the solubility of the sodium hydroxide salt) and equilibrium
$$\ce{H2O + OH- <=> OH- + H2O}$$
And what this means is that you don't have fixed hydroxil ions and water molecules, so you can't simply ignore the solvent and assume it's just a spectator molecule waiting around.
To write the two step reaction, it's not a good idea to do it in one line, because as you can see on the first step you have NaOH on both sides, which is something that doesn't make much sense. It's better to do it like this:
$$\ce{HCl + H2O -> H3O+ + Cl-}$$
$$\ce{H3O+ + Cl- + NaOH -> 2H2O + NaCl}$$
It's usually a good idea to represent salts in their ionic form when dissolved, which would be
$$\ce{H3O+ + Cl- + NaOH -> 2H2O + Na+_{(aq)} + Cl^{-}_{(aq})}$$
