I feel that as it was my comment, I am obliged to answer this :-).
First of all, birational equivalence is really a geometric notion. As far as I know, there is no analogue for groups, rings or fields and therefore the cryptographic relevance is limited. It becomes relevant when speaking of geometric objects: for example, elliptic curves.
Given these geometric objects, we want to define what it means to be "the same". The usual terminology is that given two curves $E_1$ and $E_2$, they are "the same" when they are isomorphic. There is another way to equate objects, and that is by saying that they are "almost the same". This is what a birational equivalence does: two curves $E_1$ and $E_2$ are birationally equivalent when there is a map $\phi:E_1\rightarrow E_2$ between them which is defined at every point of $E_1$ except a small set of exceptions and there is an inverse map $\phi^{-1}:E_2\rightarrow E_1$ which is defined at every point of $E_2$ except a small set of exceptions. This definition is very close to that of an isomorphism, except for the fact that we allow some exceptions.
To make this more concrete, you could think of an isomorphism as a tuple of polynomials:
$$\psi:E_1\rightarrow E_2,\quad (x,y)\mapsto (f(x,y),g(x,y)),$$
where $f,g$ are polynomials in $x,y$. The inverse is also defined in terms of polynomials.
A birational map can be thought of as a tuple of fractions of polynomials, say
$$\phi:E_1\rightarrow E_2,\quad (x,y)\mapsto \left(\frac{f_1(x,y)}{f_2(x,y)},\frac{g_1(x,y)}{g_2(x,y)}\right).$$
This is defined at all points $(x,y)$ except for the ones where $f_2(x,y)=0$ or $g_2(x,y)=0$. The inverse is also a fraction of polynomials, and can therefore be undefined at certain points.
Now let's make this even more concrete, and follow the Ed25519 paper. The Curve25519 curve is defined by $$E_1:v^2=u^3+486662u^2+u.$$
It is birationally equivalent to the Edwards curve given by
$$E_2:x^2+y^2=1+\frac{121665}{121666}x^2y^2.$$
The birational equivalence is given by the map $\phi:E_1\rightarrow E_2$ defined by
$$\phi(u,v)=\left(\frac{\sqrt{486664}u}{v},\frac{u-1}{u+1}\right).$$
Notice that it is undefined for $v=0$ or $u=-1$, and therefore is not an isomorphism. The inverse map is defined by
$$\phi^{-1}(x,y)=\left(\frac{1+y}{1-y},\frac{\sqrt{486664}u}{x}\right).$$
Again, it is undefined for $y=1$ or $x=0$.
Finally, consider the twisted Edwards curve
$$E_3:-x^2+y^2=1-\frac{121665}{121666}x^2y^2.$$
There is a map $\psi:E_2\rightarrow E_3$ defined by $\psi(x,y)=\left(ix,y\right)$, assuming that $i$ is a square root of $-1$. This is clearly defined everywhere, and is an isomorphism.
