How do I write the overall reaction for the reaction of hexaaquacobalt(II) chloride with sodium thiosulfate and ethylenediamine in the presence of #"HCl"# to form tris(ethylenediamine)cobalt(III) chloride and sodium bisulfate?

As en is a stronger-field ligand than water, the first step is fine. Then, you need to oxidize cobalt(II) to cobalt(III), and you need to provide an explanation for the source of your third chloride and extra water at the end. Let's see what happens.

Ethylenediamine has been added in this instance to exchange with the aqua ligands.

Now let's write the reduction reaction (with chloride) for it.

In an acidic environment, persulfate hydrolyzes to produce bisulfate, and cobalt(II) is expected to oxidize to cobalt(III).

After you explain how you obtained the third chloride (where else could it have come from?), the remaining sodium ions undergo the following reaction:

Let's take a look at where we are at. To account for the two tris(ethylaminediamine)cobalt(II) chlorides that we reacted with in step 3, we now need to double the amounts in the first step.

Eliminate intermediates...

and adding the sodium on both sides, I believe we've got it.

I'm not sure about the additional water in the product, but it's possible that one of the water molecules on the product side approaches, resulting in:

However, cobalt has already taken up all six of its binding sites, so it wouldn't bind.

CoCl2 · 6H2O + 2Na2S2O3 + 4en + 2HCl → [Co(en)3]Cl3 + NaHSO4 + 6H2O + 2NaCl