Intrepretation of $H^i(X, \mathcal{O}_X) = 0$ Let $X$ be a smooth proper geometrically integral variety over a field of characteristic zero (no assumptions are made on the closedness of the field). 

How should I interpret the vanishing of $H^i(X, \mathcal{O}_X)$, for, say, $i = 1,2,3$?

 A: I'll assume in order not to have complicated statements that $k$ is algebraically closed: your question is interesting enough even under that hypothesis!
Rationality
A smooth rational variety has all $H^i(X, O_X)=0$ for $i\gt0$.
So the vanishing of one of those groups is the vanishing of some obstruction to rationality.
For example:
If $X$ is a curve, rationality of $X$ is equivalent to  $H^1(X, O_X)=0$.
If $X$ is a surface Castelnuovo's criterion for rationality is the conjunction of $H^1(X, O_X)=0$ and $H^0(X, \omega^2_X)=0$. 
Topology
Over $\mathbb C$  Hodge theory implies that $H^1(X, O_X)=\frac 1 2b_1(X)$, half the first Betti number.
So the vanishing of  $H^1(X, O_X)$ is the vanishing of an obstruction to $X$ being simply connected.
For example if $X$ is a curve we have:
 $X$ simply connected $\iff$ $H^1(X,\mathcal O_X)=0$ $\iff$ $X\cong \mathbb P^1$
Picard variety
a) The tangent space at the origin of  the Picard variety  of $X$ is $H^1(X,\mathcal O_X)$.
So if $H^1(X,\mathcal O_X)=0$ the Picard group of line bundles on $X$ reduces to the Néron-Severi group and is thus a finitely generated abelian group.
For a nice example, take a smooth cubic surface in $\mathbb P^3_k$. It is rational and its Picard group is isomorphic to $\mathbb Z^7$.    
b) The condition $H^1(X, O_X)=0$ implies the nice formula valid for any variety $Y$: $$\operatorname {Pic}(X\times Y)=\operatorname {Pic}(X)\times \operatorname {Pic}(Y)      $$ Such a formula is completely false in general.
