I have heard that this set is the disjoint union of two conics in $Gr(2,4)$, but I do not have an original reference. Does anyone either have such a reference, or know a way of seeing this?
The quadric $Q$ is isomorphic to $\mathbb{P}^1 \times \mathbb{P}^1$ embedded with the linear system $\mathcal{O}_Q(1,1)$. The rulings of the quadric are the two families of lines $\mathbb{P}^1 \times \{a\}$ and $\{b\} \times \mathbb{P}^1$, and it is not difficult to see that these are the only lines in $Q$. It is also known that the locus of lines meeting a given line $\ell \subset \mathbb{P}^3$ is a hyperplane section via the Plücker embedding $G(2,4) \subset \mathbb{P}^5$ (see Harris, Algebraic Geometry, p. 244). Now by Bézout theorem a fixed line $\ell$ intersects $Q$ in two points, and for each point there is precisely a line of the first ruling and a line of the second ruling. So a hyperplane section of the Grasmannian intersects the locus corresponding to the first ruling in two points, and similarly for the second ruling. It follows that the locus corresponding to the first ruling via the Plücker embedding has degree $2$ in $\mathbb{P}^5$. On the other hand, such a locus must be a curve isomorphic to $\mathbb{P}^1$, since a ruling is a $1$dimensional linear system in $Q$; so it is a smooth conic. The same is clearly true for the locus corresponding to the second ruling. Finally, no line belongs to both rulings. Then the locus of lines of $Q$ corresponds, via the Plücker embedding $G(2,4) \subset \mathbb{P}^5$, to the disjoint union of two smooth conics. 


You can show that a smooth quadric in $\mathbb{P}^3$ is GLequivalent to the product $\mathbb{P}^1 \times \mathbb{P}^1$ embedded by the Segre map. Once you know this, you're done. (A curve on $\mathbb{P}^1 \times \mathbb{P}^1$ has a bidegree $(a,b)$, and its degree in $\mathbb{P}^3$ under the embedding $\mathbb{P}^1 \times \mathbb{P}^1 \hookrightarrow \mathbb{P}^3$ is $a+b$. Thus the lines fall in two families, corresponding to the bidegrees $(1,0)$ and $(0,1)$. ) 

