@polology:
some direct answers:
"could I think of the cocycle as the duals of cycle, so something I can integrate on the cycles, such as the forms?": qualified yes, see next answer (i.e. forms are cochains, closed forms are cocycles.)
" what are the differences between cocycles and forms?"
a real p-cochain assigns a real number to a p-chain. hence (integration over) a p-form, is a special type of p-cochain. a cocycle is a cochain with coboundary zero. hence a p-form is a cocycle if it has exterior derivative zero, i.e. a closed p-form is a particular example of a p-cocycle.
" Are the boundary relationship the same for cycle and cocycle?"
I am not sure what this is asking. if ∂ is the boundary operator on a p-chain, and if f is a (p-1) cochain, then the coboundary operator d acts on f as follows: df = f∂, i.e. since chains and cochairs are dual, the coboundary is dual to, or "adjoint" to the boundary. Since you are asking specifically about cycles and cocycles, then in a sense yes, a chain c is a cycle iff ∂c = 0, and a cochain f is a cocycle iff df = 0.
" I also read in an article that the components of this triple connection are p-form valued cochain. What does it mean?"
these are p-form valued
Cech cochains. these are dual to Cech chains. previously we have been talking about n-chains as if they were pieces of n-dimensional surfaces, i.e. simplicial or singular chains. here we are talking about Cech n-chains, which are intersections of (n+1) tuples of open sets. so a Cech n-cochain assigns a number, (or p-form) to an intersection of (n+1) open sets. this assignment is not done by integrating, but just by telling you what form goes with what intersection, e.g. va goes with Ua, and Lab goes with Uab.
there are two coboundary operators here, the Cech coboundary and the exterior derivative. hence these guys can be a cocycle in two different senses. i.e. either the Cech coboundary or the deRham exterior derivative can be zero. (These triples are made up of elements with two different indexings, the indexing by the number (n+1) of open sets, and the indexing by the degree p of the form.) the deRham coboundary is zero if the form is closed, and the Cech coboundary tells you something about compatibility of the forms on overlaps of their open sets.
e.g. a Cech 0-cochain with values in 0- forms, is a family of (smooth) functions fa, one on each open set Ua of the cover. It is a deRham cocycle iff the functions are all (locally) constant dfa = 0. It is a Cech cocycle iff the functions agree on overlaps, i.e. if the fa together define a global smooth function.
these properties are related, since if a family of functions fa are such that they differ on
overlaps by constants, i.e. if the fa are not constant but the differences fb-fa are constant, then the Cech 1- coboundary (fb-fa) on Uab is a deRham 0-cocycle, d(fb-fa)=0. It follows that the family of derivatives (dfa) defines a Cech 0-cocycle with values in 1-forms, i.e. that this family is a globally defined 1-form.
I guess the point here is that the two coboundaries apparently commute, i.e. (I am out of different letters, but if I write ∂ now for the Cech coboundary) then d∂ = 0, implies also ∂d = 0. i.e. if the Cech coboundary is deRham closed, then the deRham coboundary is well defined. this sort of argument was used in the paper to explain how to get a well defined connection 2-form (dva), from the family (va) of 1-forms. I.e. vb-va =dLab being deRham closed, implied (dva) was globally well defined.
[one way to prove the deRham theorem that topological cohomology can be computed by differential forms, is to look at such things as elements in a big double complex, and compare three different cohomologies, the deRham cohomology using d on one index and keeping the other index zero, or doing the opposite using ∂, or combining them somehow into a cohomology of the full double complex using ∂+d, then showing they are all the same. something like showing the vertical, horizontal, and diagonal cohomologies all agree. I am not expert here and have forgotten the details. this is explained in Bott-Tu.]