You have many questions, I'll try to adress them all.
A binary relation, as you read is just some set $R$ which is a subset of the cartesian product of two sets $A$ and $B$, that is $R \subseteq
A\times B$.
An example may ilustrate this:
Let $A=\{\dots,-4,-2,0,2,4,\dots \}$ (the set of even numbers), $B=\{1,3,5\}$.
Then a relation $R_1$ could be $R_1=\{(-4,1),(-4,3)(0,5)\}$
We usually denote a pair $(a,b)$ of a relation with the notation $aRb$ meaning a is related with b.
A function is a relation between two elements of two given sets condition that for each element in the domain there's one and only one image(*).
(*)That is: if $R$ is a function, $x_1\in Dom(f)$ and $y_1,y_2\in Im(f)$,
$$
x_1Ry_1 \wedge x_1Ry_2 \iff y_1=y_2
$$
Let me respond to your nitpicky question with an even more nitpicky commentary.
I assume your working definition of function is something like this:
A function is a set $f$ of ordered pairs such that if $(a, b_1) \in f$ and $(a, b_2) \in f$ then $b_1 = b_2$.
This definition allows for a function to not be one-to-one (i.e., injective): It's possible to have $(a_1, b) \in f$ and $(a_2, b) \in f$ with $a_1 \ne a_2$. However, this definition does not allow us to speak formally about a function being onto (i.e., surjective), because it does not mention the target (codomain) of the function. Here's a more thorough definition:
A function is an ordered triple $f = (A, B, \Gamma)$ where $A$ (the domain of $f$) and $B$ (the codomain of $f$) are sets, and $\Gamma$ (the graph of $f$) is a subset of the cartesian product $A \times B$ (that is, $\Gamma$ is a set of ordered pairs $(a, b)$ with $a \in A$ and $b \in B$), such that for every $a \in A$, there exists a unique $b \in B$ such that $(a, b) \in \Gamma$.
With this definition in hand, we may say that $f$ is surjective (aka. onto) if for every $b \in B$ there exists some $a \in A$ such that $(a, b) \in \Gamma$.
If $f$ is a function in the sense of definition 2, then its graph $\Gamma$ is a function in the sense of definition 1. Conversely, if $f$ is a function in the sense of definition 1, and we define $A = \{a \mid (a, b) \in f \text{ for some } b\}$ and $B = \{b \mid (a, b) \in f \text{ for some } a\}$, then $(A, B, f)$ is a surjective function in the sense of definition 2. But other choices for the codomain $B$ are possible: we can pick a bigger set $B$, and still get a definition-2 function (but it will no longer be surjective).
In other words, if we have a definition-1 function, we may "recover" its domain and its image:
$$\operatorname{dom} f := \{a \mid \exists b : (a, b) \in f\}, \qquad \operatorname{im} f := \{b \mid \exists a : (a, b) \in f\}.$$
(Please excuse the sloppy notation -- I know that applying the axiom of comprehension in this way is not quite kosher. But let's not worry about that for now.) I prefer the term image here instead of range, because range is ambiguous: It may refer to either image or codomain, depending on who you ask.
Definition 1 is preferred by books on elementary set theory. Definition 2 is more compatible with modern mathematics, so in any other context you should think of functions as objects satisfying definition 2. I like to call the former object an untyped function and the latter object a typed function: if you like, the type of $f$ is "function with domain $A$ and codomain $B$".
The definition as an ordered triple is slightly artificial, and nobody really thinks of functions this way. The important point is that the domain and codomain are actually part of the function itself, as opposed to being vague, ambient entities. Incidentally, functions may also be defined as basic "first-order" objects if you're willing to abandon standard set theory: see Tom Leinster's Rethinking set theory.
Now, to answer your original question about untyped functions: It doesn't matter, the two definitions are precisely equivalent. Because suppose $f$ is an untyped function, and $a$ is an element such that there exists $b$ with $(a, b) \in f$. Then, by definition 1, there can be no other such element $b$. In other words, $a$ is an element such that there exists a unique $b$ with $(a, b) \in f$. So the two characterizations/definitions of domain that you give are equivalent when $f$ is a function. The reason they gave the first and not the second is because the word unique is superfluous: If the word was present, it would not change anything technically, but it might be misleading because it would suggest without that word the definition wouldn't be the same. Also, when choosing between two equivalent definitions, we usually prefer the one which requires us to verify less (when e.g. we want to show that a given set is indeed the domain of a given function).
Best Answer
You are understanding one of the definitions wrong.
Surjectivity means that if you have a function $f:X\to Y$ then every $y\in Y$ gets mapped onto.
This says that something like this:
$f:\{1,2\}\to \{1,2,3\}$, $f(1)=1$, $f(1)=2$ and $f(2)=3$ is not a function, because $f(1)$ is not unique.
Other example:
Meant is something like this:
Consider the function $f:\{1,2,3\}\to\{1,2\}$
Then we can have $f(1)=1$ and $f(2)=1$ and $f(3)=2$. Then $f$ is surjective and there are values (1 and 2) that are mapped onto the same element (here 1). This is completly fine.
Keep in mind that if you have a function $f:X\to Y$. Every $x\in X$ needs to be mapped onto something. But not every $y\in Y$ needs to be mapped onto.
Example:
$g:\{1,2,3\}\to\{1,2\}$
$g(1)=g(2)=g(3)=1$ is a function. But not surjective.