What's the intuition for the fact that $\mathscr{O}(-k)$ and $\mathscr{O}(k)$ are so different? maybe this question makes no sense and I just cannot accept the fact that dual the line bundle is different from the respective line bundle itself. Since it looks like that manifolds are more intuitive than algebraic varieties, let's consider smooth complex compact manifolds.
Anyway, picking the most ordinary non-trivial line bundle over a complex manifold , namely the tautological line bundle, it's know that $0 = H^0(\mathbb{P}^n, \mathscr{O}(-k)) \not\cong H^0(\mathbb{P^n}, \mathscr{O}(k)) =$"homogeneous polynomials of degree $k$", because if both have non-trivial global sections then both must be trivial (because the manifold is compact). Another way of seeing this is by computing the cocyles of the respective bundles.
Is there an intuitive way of seeing why the above one and the dual of the above one are so different (by drawing or seeing where the glueing fails when trying to create a global section)?
Where does it fail when I try creating an isomorphism between some a line bundle and it's dual by picking fiberwise isomorphisms (as vector spaces of complex dimension $1$)?
If we just consider the smooth structure (without the holomorphic one),what happens to the global sections of both bundles in the above example? Where the fiberwise isomorphism fails to be a vector bundle (of rank $2$) isomorphism?
Thanks in advance.
 A: Since I find some comments to the question a bit ambiguous, let me emphasize:   
For $k\gt0$ the line bundles $\mathcal O(k)$ and $\mathcal O(-k)$ are not isomorphic in the $C^0$-category   and thus a fortiori not isomorphic in the $C^\infty$-category. 
In other words, the non-isomorphism you are asking about has nothing to do with algebraic geometry and can already been read on the underlying topological manifold and underlying topological bundles.
The most convincing proof is through the use of the first Chern classes:$$ c_1(\mathcal O(k))=k\neq   c_1(\mathcal O(-k))=-k\in H^2(\mathbb P^n(\mathbb C),\mathbb C)=\mathbb C$$  Beware that if a real line bundle has a riemannian metric, then it is isomorphic to its dual but if a complex line bundle $E$ has a hermitian metric (which is the case for $E=\mathcal O(k)$) you can only say that its dual $E^\ast$ is isomorphic to its conjugate bundle $\overline E$, but not to $E$ itself.  
Bibliography
By far the best resource for these results is Milnor-Stasheff's justly celebrated Characteristic Classes, especially §§ 13, 14.
