Algebraic De Rham cup product versus Betti cup product Let $X$ be a smooth projective variety over $\mathbf{C}$ of complex dimension $2n$. Let
$C_1,C_2\subseteq X$ be two closed subvarieties of complex dimension $n$. 
Then we get two Betti homology classes $[C_1],[C_2]\in H_{2n}(X,\mathbf{Z})$.
Because the singular locus of $C_1$ and $C_2$ occur (at least) in real codimension 2 we may integrate any smooth complex valued $2n$-form of $X$ on $C_i$. But more generally, if 
$\omega$ is a $\mathbf{C}$-valued real analytic $2n$-form with "no residue" (I'm not so sure of what this mean in complex dimension $>1$) it should make sense to integrate $\omega$ against $C_i$. Now let us look at the algebraic de Rham cohomology of $X$. Then $C_1$ 
and $C_2$ gives rise to two classes $C_1^*,C_2^*\in \mathbb{H}_{DR}^{2n}(X,\mathbf{C})^*$ 
(the last $*$ here means dual) which is given by integration against $C_i$. Now there is a Poincare duality which relates $\mathbb{H}_{DR}^{k}(X,\mathbf{C})$ to  $\mathbb{H}_{DR}^{4n-k}(X,\mathbf{C})$. However, beware, this Poincare duality is not induced by the integration over $X$ of the wedge product of a $k$-form with a $(4n-k)$ form.
(so all the subtlety in my question is hidden in my inability to describe with enough precision this duality). Nevertheless, from this Poincare duality we get a privileged $\mathbf{C}$-vector space isomorphism:
$$
PD:\mathbb{H}_{DR}^{2n}(X,\mathbf{C})^*\rightarrow \mathbb{H}_{DR}^{2n}(X,\mathbf{C}).
$$
Let $\iota: \mathbb{H}_{DR}^{2n}(X,\mathbf{C})\rightarrow H_{DR}^{2n}(X,\mathbf{C})$
be the comparaison isomorphism between algebraic de Rham cohomology and smooth de Rham cohomology. Define $[\eta_i]:=\iota(PD(C_i^*))$. So $\eta_i$ are smooth $2k$-forms. 
Now assume that $C_1$ and $C_2$ intersect at finitely many points. 
Q: How do we compute the period $p$ defined by the relation
$$
(\star)\;\;\;\; p\int_X \eta_1\wedge \eta_2=[C_1]\cup[C_2]\in\mathbf{Z}_{\geq 0}
$$
Note 1: In the case where $X$ is an elliptic curve and $C_1$ and $C_2$ are the two standard loops; one finds that $p=2\pi i$ (the so-called Legendre's relation). Note here tough that
$C_1$ and $C_2$ are not algebraic cycles since they have real dimension $1$. In any case, from this simple example I would expect $p$ to be a power of $2\pi i$.
Note 2: In Griffiths and Harris they prove that if $(L,\nabla)$ is a line with a connexion
on a compact Riemann surface then $c_1(L)=[\frac{-1}{2\pi i}\Theta_\nabla]$, where $c_1(L)$
corresponds to the first Chern class and $\Theta_\nabla$ corresponds to the connexion matrix of $\nabla$. One of the key idea of the computation is to apply Stoke's theorem to a differential form which has singularities. So I would like to adapt GH computation's in order to prove $(\star)$. However in order to do so I need to get a precise description of $\mathbb{H}_{DR}^{2n}(X,\mathbf{C})$ and of the map $PD$. I don't think that working with an explicit Cech-covering is the right strategy since it becomes quickly too complicated. There should be a better approach... 
 A: Since no one else answered, I'll take a stab at it, although I am not sure I'll be answering the question you are asking, since some of your notation confuses me: but here goes.
There are lots of ingredients: First, the DeRham map, defined by assigning to a differential $k$-form the cochain which  integrates the form over a smooth $k$-simplex
for any smooth manifold $M$ yields a ring isomorphism $H^*_{deRham}(M)\to H^*_{sing}(M;R)$. The ring structure is wedge product on forms and cup product on singular cohomology.  You can find proofs in Warner or Wells. Stokes' theorem is the fact that the DeRham map is a chain map from differntial forms to the singular complex.
The fact that the DeRham map is a ring map is a bit tricky, but morally follows because "what else could it be?", i.e. naturality says any natural pairing on the singular chain complex that might induce the cup product must induce the cup product on cohomology. Precise statements can be found in algebraic topology texts.
Next, given a submanifold $C^k\subset M^n$, or more generally a subvariety with singularities of real codimension at least 2, A triangulation of $C$ determines a singular $k$ cycle in $M$ denoted by $[C]\in H_k(M)$.  
(inverse) Poincare duality defines an isomorphism $P:H_k(M)\cong H^{n-k}(M)$. "Intersection theory" says that
if $C_1, C_2$ are submanifolds of $M$ of complementary dimensions ($k_1+k_2=n$), then the transverse algebraic intersection number of $C_1$ and $C_2$, $[C_1]\cdot [C_2]\in Z$ equals $(P([C_1])\cup P([C_2]))\cap [M]$, where $[M]\in H_n(M)$ denotes the fundamental class. A proof can be found in most algebraic topology books, eg Greenberg-Harper. It relies ultimately on some form  of the Thom isomorphism theorem, whose standard proof is  a spectral sequence proof. 
In your case $C_1$ and $C_2$ are  complex subvarieties, and hence every transverse intersection is positive, thus the algebraic  count (with signs) equals the geometric intersection number, and is non-negative. 
Next, "$\cap [M]$" corresponds to integrating over $M$ via the DeRham isomorphism. This follows from the identification of the cap product with the Kroneker pairing on top cohomology, and the definition of the DeRham map (e.g after triangulating $M$). Thus if $\eta_1, \eta_2$ are  closed forms whose cohomology classes are sent to $P[C_1], P[C_2]$ by the DeRham isomorphism, then using the fact that $\wedge$ maps to $\cup$, $$\int_M \eta_1\wedge \eta_2=(P[C_1]\cup P[C_2])\cap [M]=[C_1]\cdot [C_2]$$
Now, assuming this is related to what you wanted to know, the $2\pi i$ you mention are factors that come into play in Chern-Weil theory, which is the set of results that express characteristic classes of bundles over manifolds in terms of differential forms constructed out of the curvature form of any connection on the bundle. You dont need singular forms to prove these formulae: see for example Appendix C of Milnor-Stasheff for an elementary exposition.  
I dont have GH here, but presumably they express $c_1(L)$ in terms of a   divisor, i.e. the zero set of a section of $L$, and perhaps they want to insist it be holomorphic, and so the divisor might be a singular subvariety, with dual form a singular form.  
