Question regarding congruence modulo a subspace 
Definition. For $X$ a vector space, $Y$ a subspace, we say that two vectors $x_{1}, x_{2} \in X$ are congruent modulo $Y$ if $x_{1} - x_{2} \in Y$.
We can divide elements of $X$ into congruence classes mod $Y$. The congruence class containing the vector $x$ is the set of all vectors congruent with $X$; we denote it by $\{x\}$ or [$x$].

I understand the definition, but the phrasing of the following sentence is confusing to me. The definition implies that we are interested in the vectors of $X$ that are congruent to $Y$.
Could someone restate it in terms that may be more understandable?
 A: We have an equivalence relation ($xRy$ iff $x-y\in Y$) and the equivalence class of $x$ is $[x]=\{y\in X:x-y\in Y\}$.
A: Here is a more familiar example of congruence: even vs. odd integers. 
There are two equivalence classes, the set $E = \{\ldots, -4, -2, 0, 2, 4, \ldots\}$ of even integers, and the set $O = \{\ldots, -3, -1, 1, 3, \ldots\}$ of odd integers. 
Here's the cool connection: we can think of $E$ as a sort of "subspace-like" subset of the set $\Bbb Z$ of integers: it's closed under addition with other even integers, and also ("scalar") multiplication by any integers whatsoever.
So given two integers $m$ and $n$, we would say that "$m$ and $n$ are congruent modulo $E$ if $m - n \in E$"; if the difference between our two integers is even. You're probably well-familiar that the difference between two integers is even if and only if the two integers have "the same parity" (both even or both odd), so this is definitely a true statement. It's just been fancied-up to use the same kind of language you're seeing in that definition. 
The big difference is that, instead of considering a difference of integers to determine equivalence, you're using a difference of vectors. But algebraically, things look pretty much identical, since all we need is subtraction (which we can do, with integers and vectors alike).
Note: You could carry this kind of thing further, by not just considering even vs. odd, but the remainder when dividing by any integer, say $k$. You'll get $k$ equivalence classes, one for each remainder when dividing by $k$. This is all modular arithmetic is, learning to reason with equivalence classes of remainders. 
A: Here is another illuminating example in the textbook written by Peter Lax, which I found might answer your question perfectly well:
Take $X$ to be the linear space of all row vectors $(a_1,a_2,...,a_n)$ with n components, and $Y$ to be all vectors $y = (0,0,a_3,...,a_n)$ whose first two components are zero. Then two vectors are congruent iff their first two components are equal. Each equivalent class can be represented by a vector of two components, ie. the common components of all vectors in the equivalence class.
A: For easy visualization, let $X$ be $\mathbb{R}^3$ and $Y$ be a 2 dimensional subspace (i.e. a plane passing through the origin).
Now choose any vector $x_1$. Who the members of its equivalence class?
From the definition of the equivalence relation, you can construct a vector $x_2$ related to $x_1$ by
$$x_2 = x_1 + y$$
as long as $y$ is in the subspace $Y$. Do this for all the vectors $y \in Y$, and you’ve got yourself the entire equivalence class, which is nothing more than the plane $Y$ translated by the amount $x_1$. That’s why sometimes people write the equivalence class as
$$x_1 + Y$$
Incidentally, we can further equip the equivalence classes with addition and scalar multiplication in a very natural way. The result is a vector space structure, known as the quotient space.
