connections between two molecules
If the question is how to draw connections between two molecules created with separate calls of \chemfig then the answer is in principle an easy one. ChemFig's molecules are TikZ pictures, really, and the atoms are nodes in this sense. As such they can be named through ChemFig's @{<node name>} syntax and connected afterwards through some \tikz[remember picture, overlay]...; or ChemFig's wrapper \chemmove{}.
The following example
\documentclass{article}
\usepackage{chemfig}
\begin{document}
General idea:\par
\chemfig{-[7]@{a}-[1]}
\vskip2em
\chemfig{-[1]@{b}-[7]}
\chemmove{ \draw[red,dotted] (a)--(b); }
\end{document}
gives after two compilations:

In your case, however, I don't think this is the best approach. In order to be able to connect the nodes they need to have unique names. If you use one node name repeatedly only the last instance is remembered:
As a consequence the method is not so nice when the nodes to be connected are hidden in repeated submols:
\documentclass{article}
\usepackage{chemfig}
\begin{document}
problem with molecules derived from submols:\par
\definesubmol{foo}{-[7]@{a}-[1]}
\definesubmol{bar}{-[1]@{b}-[7]}
\chemfig{!{foo}!{bar}!{foo}!{bar}}
\vskip2em
\chemfig{!{foo}!{bar}!{foo}!{bar}}
\chemmove{ \draw[green,dashed] (a)--(b); }
\end{document}

Finding the right atom
This addresses your second question:
Note that the red bond (emerging from OH in C6) is not properly
placed. It should emerge from H instead than from O.
ChemFig provides a solution for this. Every bond in \chemfig{} has an optional arguments with five parameters:
-[<direction>,<length factor>,<departure>,<arrival>,<tikz options>]
The ones interesting here are <departure> and <arrival> which are integers specifying the numbers of the desired departure and arrival atoms. ChemFig's manual has the following example:
\documentclass{article}
\usepackage{chemfig}
\begin{document}
\chemfig{ABCD-[:75,,2,3]EFG}\qquad
\chemfig{ABCD-[:75,,,2]EFG}\qquad
\chemfig{ABCD-[:75,,3,2]EFG}
\end{document}

Proposed solution
In order to get two strands of cellulose I would suggest to save one submol of two glucoses with hydrogen bonds. Here's a possible definition using the turned-glucoBeta submol from my answer to the question you linked:
\definesubmol{bridged}{
-[:10,.7,,,glycosid]4
(
-[:-10](-[:150,0.7]-[2,0.7]OH)
-[:10]{\color{red}{O}}(-[:-8,1.2,,,bridge])
-[:-50,.75]
)
-[:-50]
(
-[:170]HO
-[6,,2,2,bridge]
OH
-[6,.7]
-[:-30]
-[:170]4(-[:-170,.475,,,glycosid])
-[:-50](-[:170]HO)
-[:10](-[:-50,.7]OH)
-[:-10]1(-[6]H)(-[:10,.7,,,glycosid]{\color{glyc}O}!{turned-glucoBeta})
-[:130]{\color{red}O}(-[:-8,1.2,,,bridge])
-[:-170]
)
-[:10](-[:-55,0.7]OH)
-[:-10]1(-[6,0.7]H)
-[:10,.7,,,glycosid]{\color{glyc}O}
!{turned-glucoBeta}
}
This submol can now be used repeatedly to get the polymer:

The full code
\documentclass[margin=.5cm]{standalone}
\usepackage{chemfig}
\tikzset{
bridge/.style={dotted,blue} ,
glycosid/.style={glyc}
}
\colorlet{glyc}{green}
\definesubmol{turned-glucoBeta}{
-[:-10,.7,,,glycosid]4
(
-[:10](-[:-150,0.7]-[6,0.7]OH)
-[:-10]{\color{red}{O}}-[:50,.75]
)
-[:50](-[:-170]HO)
-[:-10](-[:55,0.7]OH)
-[:10]1(-[2,0.7]H)
-[:-10,.7,,,glycosid]{\color{glyc}{O}}
}
\definesubmol{bridged}{
-[:10,.7,,,glycosid]4
(
-[:-10](-[:150,0.7]-[2,0.7]OH)
-[:10]{\color{red}{O}}(-[:-8,1.2,,,bridge])
-[:-50,.75]
)
-[:-50]
(
-[:170]HO
-[6,,2,2,bridge]
OH
-[6,.7]
-[:-30]
-[:170]4(-[:-170,.475,,,glycosid])
-[:-50](-[:170]HO)
-[:10](-[:-50,.7]OH)
-[:-10]1(-[6]H)(-[:10,.7,,,glycosid]{\color{glyc}O}!{turned-glucoBeta})
-[:130]{\color{red}O}(-[:-8,1.2,,,bridge])
-[:-170]
)
-[:10](-[:-55,0.7]OH)
-[:-10]1(-[6,0.7]H)
-[:10,.7,,,glycosid]{\color{glyc}O}
!{turned-glucoBeta}
}
\begin{document}
\chemfig{!{bridged}!{bridged}!{bridged}}
\end{document}