You are right, the first chern class of a principal divisor is $0$: in fact this statement lies in the definition of "divisor".
You can think as Chern classes as cycles (that is, the generalization of divisors to higher codimension) rather than cohomology classes. So you can think that Chern classes associate to a vector bundle $E$ on $X$ a cycle $c_i(X)$.
Considering the space of cycles is not relevant enough: you quotient your space of cycles by rational equivalence (that is, two cycles are equivalent if they differ from a "principal" cycle). You obtain a nice space called Chow group $CH^*(X)$, see https://en.wikipedia.org/wiki/Chow_group.
In the particular case of line bundles, the first Chern class associate to a line bundle $L$ the (equivalent class of) the divisor given by the zeros of some generic global section. Starting from a divisor $D$, you can define a line bundle $L(D)$: in fact you just go the opposite direction. In terms of cycle, you have $c_1(L(D))\equiv D$.
Finally, you can go from the Chow group to the cohomology of your variety $H^{2*}(X,\mathbb{Q})$ using Poincaré duality (equivalent cycles have same image in cohomology!), that's why we often say $c_1(L)\in H^2(X,\mathbb{Z})$. Principal divisor are already $0$ in the Chow group, thus so are their images in cohomology.
Just to get this off the unanswered list.
As mentioned in the original post, an effective divisor $D$ on $\mathbb{P}^n$ is a finite sum $\displaystyle \sum_i a_i V_i$ where $a_i\in\mathbb{Z}^{\geqslant 0}$ and $V_i$ are irreducible subvarieties of $\mathbb{P}^n$. We call such a divisor positive if the associated line bundle $\mathcal{O}(D)$ has a Hermitian metric with positive curvature. It's easy to see that if $D_1,\ldots,D_m$ are positive then so is $D_1+\cdots +D_m$.
So, to see that every effective divisor on $\mathbb{P}^n$ is positive it suffices to show that every divisor of the form $V$ is positive where $V$ is a closed analytic submanifold. But, by Chow's theorem we know that $V=V(f)$ where $f$ is a homogenous polynomial in $n+1$-variables of degree $d$. But, one can quickly check by hand that $V(f)$ is then equivalent to $dH$ where
$$\mathbb{P}^{n-1}\cong H:=\{[0:z_1:\cdots:z_n]\}\subseteq \mathbb{P}^n$$
So, by our above observations it suffices to show that $H$ is positive. But, this is clear.
One can also verify the claim that every $V$ is linearly equivalent to a multiple of $H$ without appealing to Chow's theorem as follows. We know that the line bundles on $\mathbb{P}^n$ are classified by $H^1(\mathbb{P}^n,\mathcal{O}^\times)$. But, by the exponential sequence
$$0\to \underline{2\pi i\mathbb{Z}}\to \mathcal{O}\to \mathcal{O}^\times\to 0$$
we get the exact sequence
$$H^1(\mathbb{P}^n,\underline{2\pi i\mathbb{Z}})\to H^1(\mathbb{P}^n,\mathcal{O})\to H^1(\mathbb{P}^n,\mathcal{O}^\times)\to H^2(\mathbb{P}^n,\underline{2\pi i \mathbb{Z}})\to H^2(\mathbb{P}^n,\mathcal{O})$$
But,
$$H^i(\mathbb{P}^n,\underline{2\pi i \mathbb{Z}})\cong H^i_\text{sing}(\mathbb{P}^n,\mathbb{Z})=\begin{cases} \mathbb{Z} & \mbox{if}\quad i\in\{0,2,\ldots,2n\}\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$$
and
$$H^i(\mathbb{P}^n,\mathcal{O})=\begin{cases} \mathbb{C} & \mbox{if}\quad i=0\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$$
from where we see that
$$\mathrm{Pic}(\mathbb{P}^n)\cong H^1(\mathbb{P}^n,\mathcal{O}^\times)\cong H^1(\mathbb{P}^n,\underline{2\pi i \mathbb{Z}})\cong \mathbb{Z}$$
call the map
$$H^1(\mathbb{P}^n,\mathcal{O}^\times)\to H^2(\mathbb{P}^n,\underline{2\pi i\mathbb{Z}})\cong \mathbb{Z}$$
the 'Chern class map' and denote it by $c$. What the above shows is that a line bundle on $\mathbb{P}^n$ is entirely determined by its Chern class.
Now, it's not hard to directly check that $c(\mathcal{O}(H))=1$ and in general $c(\mathcal{O}(D))>0$ if $D$ is effective (just think about the cycle class map) from which case it follows that $\mathcal{O}(D)\cong \mathcal{O}(H)^{c(\mathcal{O}(D))}$ or, in other words, $D\sim c(\mathcal{O}(D))H$ (where $\sim$ denotes equivalence).
Best Answer
This is all correct!
Generally speaking, there's a mismatch with the definitions some people immediately latch to when hearing/using the words 'line-bundle'. The 'algebraic geometry' way to define a line bundle $\mathcal{L}$ on a variety $X/\mathbf{C}$ is as a sheaf of $\mathcal{O}_X$-modules which is locally (either in $X$'s analytic or Zariski topology, if it's algebraic - this ends up not mattering if $X$ is compact for instance by a theorem of Serre) isomorphic to the free rank-one $\mathcal{O}_X$-module. From such sheaves one can then contruct (very algebraically) an actual line bundle $L \to X$ called the 'geometric vector bundle associated to the sheaf $\mathcal{L}$' and this produces an equivalence between the two notions. Moreover and crucially for your question, because this equivalence is functorial in $X$, any open subset on which the sheaf $\mathcal{L}$ is isomorphic to $\mathcal{O}_X$ is also an open on which the line bundle $L \to X$ restrict to a product of a one-dimentional vector space with the base and viceversa: the gluing data for $L$ and $\mathcal{L}$ provide the same Cech cocycle in the sheaf cohomology group $H^1(X,\mathcal{O}_X)$ and thus the abuse of notation $c_1(L) = c_1(\mathcal{L})$ is justified.