A question related to the concept of being "relatively prime" Suppose that I have $a, b, c, d \in \mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers.
If I have the equation $ab = 2cd$ and I know that $\gcd(a,c)=\gcd(c,d)=1$, then it follows that I have the following divisibility constraints:
$$a \mid 2d,$$
and
$$c \mid b.$$
Is this correct?

Your results are correct (but the second assumption $\gcd(c,d)=1$ is not used), For this sort of problems, the theorem to keep in mind is Euler's Lemma:
$$\left(x|yz \text{ and } \gcd(x,y)=1\right)\Rightarrow x|z \tag 1$$
Now return to your question $ab=2cd$,using $(1)$ for $x=a$, $y=c$ and $z=2d$  we conclude that $a|2d$ so we can write $2d=qa$ and hence 
$$b=qc$$
and from here follows that $c|b$; 
