0. On the "finite dimension": By all means, that is just a glitch. When you scroll down in the source (and many others), you will see that most of the time we only look at Lie algebras which are of finite dimension over their ground field. Milne notes that a few lines later. Maybe he originally had put the assumption of finite dimension in the definition of a Lie algebra (some sources do that for convenience), and then any abelian Lie algebra would be finite-dimensional, not because it's abelian, but because it is a Lie algebra. But now the definition of Lie algebras does not include finite dimension, and yes, being abelian or commutative does not imply anything about the dimension, so one should just delete "finite-dimensional" in that line.
1. Let's spell out some very basic linear algebra which Randall points out in comments but which seems to cause problems for OP:
Lemma. Let $V$ be any vector space over a field $K$. If $\mathrm{char}(K) \nmid n$, then $\underbrace{v + v + \dots+ v}_{n \text{ times}} = 0$ implies $v=0$.
Proof as one would normally write it: $\mathrm{char}(K) \nmid n \implies \frac{1}{n} \in K$ and hence $nv=0$ implies $v = \frac{1}{n} \cdot nv =0$.
Ultra-formal proof: In the following, $+_K$ and $\cdot_K$ mean addition and multiplication in the field $K$, while $+$ means addition in the vector space $V$ and $\cdot$ means multiplication of a scalar from $K$ with a vector from $V$. Further, $1$ means the multiplicative identity in $K$, $0_K$ means the additive identity in $K$, while $0$ means the additive identity ("zero vector") in $V$.
Now, $\mathrm{char}(K) \nmid n$ implies that $n_K := \underbrace{(1 +_K 1 +_K \dots +_K 1)}_{n \text{ times}} \in K \setminus \{0_K\}$, so it has a multiplicative inverse $n^{-1}_K \in K$, and hence
$$v = 1 \cdot v = (n^{-1}_K \cdot_K n_K) \cdot v = n^{-1}_K \cdot (n_K \cdot v) \\= n^{-1}_K \cdot \left(\underbrace{(1 +_K 1 +_K \dots +_K 1)}_{n \text{ times}} \cdot v\right) \\= n^{-1}_K \cdot \underbrace{(1\cdot v + 1\cdot v + \dots +1\cdot v)}_{n \text{ times}} \\= n^{-1}_K \cdot \underbrace{(v + v + \dots+ v)}_{n \text{ times}} \\ \stackrel{(*)}= n^{-1}_K \cdot 0 = 0$$
where $(*)$ was our assumption, and each other equality is some definition or axiom of what a vector space over a field is (the fifth equal sign in particular is what Randall calls "the distribution axiom").
2. A note on characteristic $2$ and on terminology:
Besides bilinearity and the Jacobi identity, the Lie bracket in a Lie algebra $L$ satisfies one further axiom. The proper way to state this axiom is
$$\forall x \in L: [x,x] = 0 \qquad (A1)$$
Milne states it like this too. Some sources instead give the axiom
$$\forall x,y \in L: [x,y] = -[y,x] \qquad (A2)$$
As Milne notes, and please solve for yourself, it is an easy linear algebra exercise (using bilinearity of $[,]$) that $A1 \implies A2$ regardless of the characteristic of $K$. You can also find an easy proof that $A2 \implies A1$ but only if at some point you assume $\mathrm{char}(K) \neq 2$. This is the reason why $(A1)$ is preferred. Surely sources where the ground field is never anything but $\mathbb R$ or $\mathbb C$ understandably don't care much. But good sources (like, in this case, Wikipedia) point out the subtle difference.
Note that accordingly, there is a slight deviation of terminology in characteristic $2$. The consensus right now seems to be that a Lie algebra in which
$$\forall x,y \in L: [x,y] = 0 $$
is called "abelian", while a Lie algebra in which
$$\forall x,y \in L: [x,y]=[y,x]$$
is called "commutative". With this terminology, it is obvious that abelian implies commutative, and the whole point of your first question and my ultra-formal paragraph 1 was to convince you that in case $\mathrm{char}(K) \neq 2$, commutative implies abelian i.e. they are the same thing. In characteristic $2$ however, with this terminology, either of $A1$ or $A2$ implies that every Lie algebra is commutative (but not necessarily abelian). Cf. Dietrich Burde's answer here.
3. Why do we say "$x$ and $y$ commute" if $[x,y]=0$:
This is explained well in the other answer. Confer also e.g. Why do we use the commutator bracket for Lie algebra's. In standard (and historically first) examples, the Lie bracket was defined as a new operator on some associative algebra (matrices or vector fields) via $$[X,Y] := XY-YX.$$ The thing on the right hand side obviously measures "how much $X$ and $Y$ fail to commute" in the associative algebra setting, which is why it is called "the commutator". To say "the commutator is zero" is then just a different way of saying that $X$ and $Y$ commute in the original associative algebra. --- Now, not all Lie algebras "really come from" associative algebras like that (although one can try to "envelop" them with such), but the terminology is still handy and widely used.
Note that according to what's written above in 2, in all characteristics except $2$, we have that $[x,y]= 0$ is equivalent to $[x,y] =[y,x]$. Whereas in characteristic $2$, there is a difference, namely the Lie bracket can (actually, always does) "commute" even though when viewing the bracket as commutator of an associative algebra, the elements do not commute in that associative algebra. (In a way, commutativity of the commutator becomes a different concept than commutativity of the "original associative" elements, whereas in every other characteristic that is just the same.)
4. "To give an abelian Lie algebra is to give a vector space". (With all said above, it should be clear now that this is how the sentence should be written.)
You are not asking about this, but I just want to prepare you that at some point, somebody might phrase that as "The forgetful functor from the category of $K$-Lie algebras to the category of $K$-vector spaces, when restricted to the subcategory of abelian Lie algebras, gives an equivalence of categories". What this very fancy terminology means is that, if you have any $K$-vector space $V$, there is one and only one way to "make it" into an abelian Lie algebra, namely, by defining $[x,y] := 0 $ for all $x,y \in V$; and further, if you have two $K$-vector spaces and a $K$-linear map between them $f: V \rightarrow W$, then if you turn $V, W$ into abelian Lie algebras via that one and only and obvious way, then that map $f$ is now also a homomorphism of Lie algebras. (And if $f$ was an isomorphism of vector spaces, then now it is an isomorphism of Lie algebras.)
The idea behind this, again well explained in the other answer too, is that if you have an abelian Lie algebra, you cannot "know more" about it than whatever you know about it as a vector space. If you want, for an abelian Lie algebra you can just "forget" that it is a Lie algebra at all and just treat it as a vector space: you do not really lose information, because you can always "turn it back" into an abelian Lie algebra in one and only one (and obvious) way.
