Let me run through it again, but I'll do 2 things. I'll answer your specific questions in the early part of the proof, and I'll take a slightly different tact for the latter part to see if it makes more sense to you. Bear with me if I go into details on stuff you already understand.
Proof: Assume that $\sqrt{2}$ is rational, then $\sqrt{2}= {p\over q}$ for some integer $p$ and $q$ with $q≠0$ and gcd($p$,$q$)=$1$.
(2 things to note here: First, the only thing we assumed is that $\sqrt{2}$ is rational. Everything else above follows directly from that assumption and the definition of rational.
Second, gcd = $1$ means no integer greater than 1 divides both $p$ and $q$. More specifically, $2$ cannot divide them both, so they cannot both be even.)
$(p/q)^ 2 =2={{p^ 2}\over {q^2}}$
(What was the purpose of squaring √2 and p/q?) To get rid of the square root.
$p^2=2q^2$
(Is this basically just canceling out the q^2 on left side, and multiplying the right side?) Yes, to get rid of the fractions.
(Now, here's where I veer off.)
By that last equation, $p^2$ is even (because it's $2$ times something). If $p$ itself were an odd number, then $p^2$ would be odd, so we know now that p is Even.
Now, let's divide both sides by $2$.
${{p^2}\over 2} =q^2$, rearranging a bit, we get
${p\over 2}\times p =q^2$
Now, $p$ is even, so ${p\over 2}$ is an integer. And since $p$ is even, that integer times $p$ is even.
So ${p\over 2}\times p$ is even. Which means $q^2$ is even. But by the same logic we used above, if $q$ itself were an odd number, then $q^2$ would be odd, so we know now that q is Even.
But we said at the beginning that this cannot be true, so we have a contradiction. And since we do, it means one of our assumptions is false. But we only made one assumption, so $\sqrt 2$ must not be rational.
(Final note: all that $2k$ stuff in the original proof was one way of demonstrating that there are two $2$'s in the $p^2$, and so when you divide both sides by $2$, what remains on the left is still even. However, sometimes adding new variables throws people, so I went a different way. Hope it helped.)