Appending multiple non-nested elements for each data member with D3.js

Use append() for the first item and insert() for the second. This eliminates any need to sort afterwards (thanks to @scuerda's comment for pointing this out) (JSFiddle):

data = [{}, {}, {}];
var enterSelection = d3.select('#parent').selectAll('p').data(data).enter()

enterSelection.append('p').text(function(d, i) {return 'from data[' + i + ']'})
enterSelection.insert('p').text(function(d, i) {return 'from data[' + i + ']'})

This will give you the exact structure requested:

<p>from data[0]</p>

<p>from data[0]</p>

<p>from data[1]</p>

<p>from data[1]</p>

<p>from data[2]</p>

<p>from data[2]</p>

The idomatic way of doing is with nesting:

var divs = d3.select('#parent').selectAll('p').data(data).enter().append('div');

divs.append('p')
divs.append('p')

Which creates:

<div id="parent">
  <div>
    <p> from data[0] </p>
    <p> from data[0] </p>
  </div>

  <div>
    <p> from data[1] </p>
    <p> from data[1] </p>
  </div>

  <div>
    <p> from data[2] </p>
    <p> from data[2] </p>
  </div>
</div>

If that won't work, save your selection and repeatedly append:

var enterSelection = d3.select('#parent').selectAll('p').data(data).enter();

enterSelection.append('p')
enterSelection.append('p')

then sort what you've added:

d3.select('#parent').selectAll('p').sort(function(a, b){ return a.index - b.index; })

You'll need to add an index property to each element of data that describes the sort order. The normal i is only defined in the context of a particular selection, which is lost when we reselect.