(Throughout this post, I am talking about vector spaces.)
I had the pleasure of doing Abstract Algebra two semesters early, however, I feel like some general context was lost in the process. While I have a certain intuitive understanding of both quotient groups and cosets, I had never heard the terms "quotient space" or "equivalence classes". I am trying to create an intuitive notion of quotient spaces, based on my knowledge of quotient groups.
Consider the group $\mathbf{Z}\times\mathbf{Z}$, and its normal subgroup $\langle(1,0)\rangle$. We see that the quotient group $G/N$ is isomorphic to $\mathbf{Z}$. In other words, it is still a group. In particular, we can read this as $G \:\text{mod}\:N$, which gives an even more intuitive notion of a quotient groups. (And helps a lot when working with polynomial rings.)
Do quotient spaces and equivalence classes follow in an obvious way? Is there an example, of the same simplicity as the example provided above that can be used to explain the connection?
The short answer is that, as vector spaces are a particular type of group (namely, abelian groups) their quotients are particular examples of group quotients. In fact, much of the technical detail of the quotient construction is not apparent for vector spaces, since as abelian groups, all their subgroups are normal. There is, however, the additional technical detail that a general sub group of a vector space may not be a sub space. If you take the quotient of a vector space by a vector subspace, it is also a vector space; otherwise, it is just an abelian group.
If you understand the concept of $\def\Z{\mathbb{Z}}\def\genby#1{\langle#1\rangle}(\Z \times \Z)/\genby{(1,0)}$, and if you also understand some computational methods in linear algebra, then you basically understand the mechanics of taking (finite dimensional) vector space quotients. Indeed, to compute $V/W$, where $W$ is any subspace of $V$, pick a basis for $W$ and extend it to a larger basis for $V$. Then we have $\def\R{\mathbb{R}}V \cong \R^n$ and $W = \genby{(1,0,\dots, 0), \dots, (0, \dots, 1, \dots, 0)} \cong \R^m$ (it is generated by some number of the basis vectors for $V$). The quotient $$\R^n/\genby{(1,0,\dots, 0), \dots, (0, \dots, 1, \dots, 0)} \cong \R^{n - m}$$ corresponds to the basis vectors you didn't put in $W$.
There is a significant issue here, which is that it looks like this makes the quotient a subspace of $V$, which is actually not the case. What's true is that, as a result of this computation, we have represented it as a subspace, but with a different choice of basis vectors, we would have a different representation. Naturally, $V/W$ is an abstract vector space.
In order to describe this, indeed, the general language of quotients that you know from group theory applies to vector spaces as well. The quotient $V/W$ is constructed as the set of cosets $v + W$ for $v \in V$, given the natural operations (addition and, since we are talking about vectors, scalar multiplication act on the $v$ part, which is well-defined). This partitions $V$ into equivalence classes (i.e. the relation $v \, R \, w$ iff $v - w \in W$ is an equivalence relation on $V$). And most importantly, $V/W$ satisfies the isomorphism theorem: there is a "quotient map" $V \to V/W$ sending $v$ to $v + W$, and if $U$ is any vector space and $T \colon V \to U$ is any linear map such that $T(W) = 0$, then there is a unique linear map $V/W \to U$ whose composition with the quotient map is $T$. You will find, as you travel through life, that this is the most essential aspect of quotients.