If I have u = ( u 1 , u 2 ) is a vector...

The curl of a vector field is only really defined on vector fields $\mathbf{F}:{\mathbb{R}}^3\to {\mathbb{R}}^3$. For a two-dimensional field $u=(u_1(x,y),u_2(x,y))$the z component is taken to be zero. The resulting curl has non-zero components only in the z-direction:
$\mathrm{\nabla }\times (u_1,u_2,0)=(0,0,\frac{\mathrm{\partial }{u}_2}{\mathrm{\partial }x}-\frac{\mathrm{\partial }{u}_1}{\mathrm{\partial }y})$
so people sometimes call the following scalar quantity the "curl":
$\frac{\mathrm{\partial }{u}_2}{\mathrm{\partial }x}-\frac{\mathrm{\partial }{u}_1}{\mathrm{\partial }y}$

The curl of a 2D vector can be defined with the following "intuition": The 2D vector $u=(u_1,u_2)$ can be "extended" to ${\mathbb{R}}^3$ as $\hat{u}=(u_1,u_2,0)$. Now we can take the "usual" curl of this $\hat{u}$ vector: $\hat{v}:=\mathrm{\nabla }\times \hat{u}=(0,0,v_3)$ and we identify $v_3$ with the curl of u, i.e.
$\text{curl}(u_1,u_2)={\epsilon }_{321}{\mathrm{\partial }}_2{u}_1+{\epsilon }_{312}{\mathrm{\partial }}_1{u}_2=\frac{\mathrm{\partial }{u}_2}{\mathrm{\partial }x}-\frac{\mathrm{\partial }{u}_1}{\mathrm{\partial }y}$