Let $Gr$ denote the Grassmannian and let $Gr(2, T\mathbb R^3) = \bigcup_{x \in T\mathbb R^3} Gr(2, T_x \mathbb R^3)$.
Consider one $2$-dimensional subspace of $\mathbb R^3$, that is, one element of $Gr(2, T_x \mathbb R^3)$ for some $x$.
Then if $\{(1,0,a),(0,1,b)\}$ is a basis of this $2$-subspace, $(a,b)$ is a chart on this subspace. (see here)
But I don't understand why this should be true.
Please could someone explain to me, in as simple as possible words, why the last coordinates of two basis vectors give a chart for the subspace?
Consider the $xy$-plane $L_0$ as an element in $Gr(2, 3)$ (the set of all two planes in $\mathbb R^3$). Let $U$ be the subset in $Gr(2, 3)$, such that $L \in U$ if and only if $L$ does not contain the $z$-axis. Note that this is the same as saying that (check)
$L \in U$ if and only if $\pi_L:= \pi|_L : L \to L_0$ is an isomorphism, where $\pi : \mathbb R^3 \to \mathbb R^2 \cong L_0$ is the projection to the $xy$-plane.
Now note that $\pi_L^{-1}$ as a linear map is completely determined by its action on $(1,0,0)$ and $(0,1,0)$. Let $(a, b)$ be given by
$$\pi_L^{-1} (1,0,0) = (1, 0, a),\ \ \pi_L^{-1} (0,1,0) = (0, 1, b)$$
and define $\phi : U \to \mathbb R^2$ be given by $\phi(L) = (a, b)$. Then $\phi$ is a bijection and so define a chart on $Gr(2, 3)$.