Let $L$ be a one-dimensional subspace of $\mathbb{R}^2$, namely, a line, say, the set of $(x,y)$ with $y = x$. and consider the quotient $\mathbb{R}^2/L$. The equivalence classes, which make up the elements of this quotient space, are exactly the set of ordered pairs in $\mathbb{R}^2$ whose difference is an element of the line $y = x$. The equivalence class of the point $(a,b)$, assuming $a \neq b$, ends up being the line through $(a,b)$ parallel to the line $y = x$.
This geometry makes sense to me. By simple number-crunching, I can also see that $$\dim \mathbb{R}^2/L = \dim \mathbb{R}^2 - \dim L = 2 - 1 = 1.$$ I am trying to understand why, geometrically, it makes intuitive sense for the dimension to be $1$ or why it makes sense to say there is "one degree of freedom." When I sketch the picture, it almost looks like I'm picking a line up and "scaling it" across the $\mathbb{R}^2$ plane, in which case the only degree of freedom is "vertical distance" of some kind. Is there a better way to think of this?
Since you quotient by a line, it can be convenient to choose your coordinates so that one axis, say the $y$-axis is coincident with $L$. Then every point has coordinates in this coordinate system and the quotient collapses each point $(x,y) \mapsto x$. That is, the quotient acts like the linear map $(x,y) \mapsto 1 \cdot x + 0 \cdot y$. Now look at the nullspace of this map, which is one dimensional and think about the rank-nullity theorem.