Question: Suppose I know the graphs of two functions $f(x)$ and $g(x)$. How can I find the graph of $h(x)=f(x)+g(x)$? What are the rules to be followed ? 

P.S. In case my question seems silly,at least provide me with a link or something so that I can learn!

Answer: 
Hodge theory is a vast subject, especially since Deligne and Griffiths revolutionized it and algebraized it around 1970 with the introduction of variations of Hodge structures and mixed Hodge structures, axiomatic approaches which found amazing applications in algebraic geometry. 

At the most elementary level I would say that Hodge theory is a refinement of De Rham cohomology:  

De Rham proved that classical singular cohomology $H^\*\_{sing}(M,\mathbb R)$ of a smooth compact manifold $M$ is the same as the cohomology $H^\*\_{deRham}(M)$ of the complex of smooth differential forms $\Omega^\*(M)$.  

This means that a cohomology class can be represented by a closed differential form.  

 But which one should one choose? Unfortunately there is no canonical choice.  

Hodge theory furnishes a nice solution to this query: introduce a Riemannian structure on $M$ and then every cohomology class $c\in H^k\_{deRham}(M)$ will contain one and ond only one closed harmonic form $\omega \in \Omega^k\_{Harm}(M)$ representing it : $c=[\omega]$.  

Harmonic here means annihilated by the Laplace-Beltrami operator $\Delta^k:\Omega^k(M)\to \Omega^k(M)$, an analogue for $k$-forms of the standard Laplace operator $\Delta=\Delta^0:C^\infty(M) \to C^\infty(M)$ for functions. 

Hodge theory immediately allows you to prove the finite-dimensionality of $H^k\_{deRham}(M)$ and also gets you Poincaré duality for free: $H^{n-p}\_{deRham}(M)\simeq (H^{p}\_{deRham}(M))^\*$  

The price to pay for Hodge theory is, however, some investment in the theory of elliptic operators, which some geometers ignorant of partial differential equations (like me) find hard going.
I would say that the basic intuition is that a Riemannian metric defines an inner product on the space of differential forms, which can be used to define an adjoint operator to the exterior derivative.

Now consider the adjoint $(d\_{k-1})^\*:\Omega^k\to\Omega^{k-1}$ of $d\_{k-1}:\Omega^{k-1}\to\Omega^k$. **If the spaces $\Omega^\ell$ were all finite dimensional**, then $\ker((d\_{k-1})^\*)$ would be the orthocomplement of $\DeclareMathOperator{\im}{im}\im(d\_{k-1})$. Hence $\ker((d\_{k-1})^\*)\cap \ker(d\_k)$ would be a complement to $\im(d\_{k-1})$ in $\ker(d\_k)$ and thus project isomorphically onto the cohomology space $H^k$. Hence you see that $\ker((d\_{k-1})^\*)\cap \ker(d\_k)$ would have the same dimension as $H^k$.

Using adjointness again, it is easy to verify that (if things were finite dimensional) defining $\Delta\_k=d\_{k-1}\circ (d\_{k-1})^\*+(d\_k)^\*\circ d\_k$ you get $\ker(\Delta\_k)+\ker((d\_{k-1})^\*)\cap \ker(d\_k)$. (This is like $\Delta(\phi)=0$ implies $0=d\Delta(\phi)=dd^\*d\phi=0$ (since $dd=0$) and using adjointness twice, one gets $d\phi=0$, and likewise for $d^\*$.)

Using functional analysis/PDE theory as mentioned in the answer of @GeorgesElencwajg , one proves that this continues to hold on a compact manifold in spite of the fact that one deals with infinite dimensional spaces.