Determine if $y = x^2$ is injective I realize that $y=x^2$ is not injective. It is not one-to-one ($1$ and $-1$ both map to 1, for example). 

However, in class it was stated that a function is injective if $f(x) = f(y)$ implies $x = y$.

Or if $x$ doesn't equal $y$, then this implies that $f(x)$ doesn't equal $f(y)$.

This is where I'm confused. (Or maybe tired.) For $x = 2$, $y = 4$. So, $f(x) = 4$, but $f(y) = 2$ ($\sqrt{y} = x$). Therefore, $x$ and $y$ are not equal, so it's not injective.

However, according to the contrapositive, $x$ doesn't equal $y$ implies that $f(x)$ doesn't equal $f(y)$. This fits.

Do both the contrapositive and the contrapositive of the contrapositive have to be true for it to be injective? Or am I doing something stupid?
The statement in class is correct, and you example of $x=1, y=-1$ proves the function is not injective because you have $f(x)=f(y)$ but $x \neq y$.  The contrapositive fails as well because you have $x \neq y$ but $f(x)=f(y)$ The statement and its contrapositive are logically equivalent, so you only need to check one of them.
I think that the syntax of the definition from your class is the point of confusion. The class definition depends on functions being defined as 
$$
f(x)=(\text{expression of }x)
$$ rather than as 
$$
y=(\text{expression of }x).
$$ Consequently, the "$y$" in "$f(y)$" is just some dummy variable that gets input into $f$ and is unrelated to $x$. So, for your example, $f(x)=x^2$, $f(x)= 4$ for $x=2$ and $f(y)=f(4)=16$ for $y=4$. Again, note that $y$ is unrelated to $x$.

Finally, in order to show that $f(x)=x^2$ isn't injective, you can start with the definition or with its contrapositive, as you stated:


Starting with the definition, 
$$
f(-2)=f(2)\:\text{ but }-2\:\text{ isn't equal to }2.$$ 
Starting with the contrapositive, let's consider $x=2$ and $y=-2$: then 
$$
x\neq y\:\text{ but }\:f(x)=f(y)=4.
$$