Two varieties $X,Y$ are said to be birational if there exist rational maps in each direction such that either composition is the identity on a open dense subset. Note that here the morphisms aren't assumed to be over any base scheme.
However, to say that an integral finite type $k$-scheme is rational (i.e. birational to $\Bbb A_k^n$), one imposes that the rational map (and its inverse) to $\Bbb A_k^n$ be over $k$. Why?
An integral finite type $k$-scheme, by definition, can be covered by finitely many $U_i =\operatorname{Spec}k[x_1,...,x_{n_i}]/I_i$, but, if I understood everything correctly so far, there is no restriction on how $k$ maps into $k[x_1,...,x_{n_i}]/I_i.$ I suppose that it doesn't even have to land in $k$ as shows the example $\phi:\Bbb Q(\sqrt 2)\rightarrow\Bbb Q(\sqrt 2)[x_1]/(x_1^2-2)$ by $1\mapsto 1, \sqrt2\mapsto x_1$.
EDIT: I realize now that what I said in the last paragraph is not entirely correct. For a finite type $k$-scheme, one can actually find affine opens $\operatorname{Spec} B_i$ such that $B_i$ is a finitely generated $k$-algebra. This means that $k$ maps into $B_i=k[x_1,...,x_{n_i}]/I$ in the obvious (standard) way. This doesn't mean, however, that you cannot find other $A_j$, also of the form $k[x_1,...,x_{n_j}]/J$ into which the map induced by $\operatorname{Spec}k\rightarrow X$ maps $k$ into $A_j$ in a non-standard way (as in my example above).
Which source are you reading? When dealing with varieties over a field $k$, of course one only considers morphisms over $k$. This applies - in particular - to birational maps. They are (partially defined) $k$-morphisms.
When you consider morphisms not over $k$, you actually leave the category of varieties and work in the larger category of schemes. So we have a(nother) secret application of a forgetful functor.