Field of formal rational function and $K(x_1, \dots, x_n)$ Def.  Let $K\subseteq L$ be a field extension, and let $B$ be a subset of $L$. We denote by $K(B)$ the smallest subfield of $L$ containing both $K$ and $B$.
Now, let $R=K[x_1,\dots, x_n]$ be the polynomial ring in $n$ variables over a field $K$, and let $$L=\text{Quot}R=\bigg\{\frac{f}{g}\;:\;f, g\in R, g\ne 0\bigg\}$$ be its quotient field, i.e the field of formal rational functions in $n$ variables over $K$.
This field is usually denoted by $K(x_1\dots, x_n)$, and thus by the same round bracket notation as for the smallest extension field of $K$ containing given elements of a larger field, which can be descrived explicity as the set of all rational functions in these elements with coefficients in $K$.

Question Is this ambiguity just notational or are they actually the same field?

 A: Let me repeat what I said in the comments. There is  no ambiguity here. We have the standard notation for the field of fractions $K(x_1,x_2,\dots,x_n)$, and we have a definition of what $K(B)$ means for a set.
The question remains whether $K(x_1,x_2,\dots,x_n)=K(\{x_1,x_2,\dots,x_n\})$. The answer is "of course" and here are some of the details.
Let $K$ be a field, $R:=K[x_1,x_2,\dots, x_n]$ be the polynomial ring in $n$ variables.
Let $L$ denote the field of fractions of $R$, so that $L=\left\{\frac{f}{g}\mid f,g\in R, g\ne0\right\}$. This field is usually denoted by $K(x_1,x_2,\dots,x_n)$.
Now let $B$ be the set $\{x_1,x_2,\dots,x_n\}$. Then consider the field $K(B)$ defined as the smallest subfield of $L$ containing $K$ and the set $B$.
By definition we have that $K(B)\subseteq L$.
Now let $\frac{f}{g}\in L$. We have that $f,g\in K[x_1,x_2,\dots,x_n]$. We can write $f=\sum_{} a_{i_1,i_2,\dots,i_n} x_1^{i_1}x_2^{i_2}\dots x_n^{i_n}$ where the ceofficients are in $K$ and the sum is finite. Now $K(B)$ by definition contains all the $x_i$ and so (being closed under multiplication) contains all the monomials $x_1^{i_1}x_2^{i_2}\dots x_n^{i_n}$. $K(B)$ also by definition contains all the coefficients as these are in $K$. Hence, as $K(B)$ is a field it must contain $f$. The same is true for $g$. But $K(B)$ is a field so it must contain also $\frac{f}{g}$. That is $L\subseteq K(B)$.
We therefore have that $L=K(B)$.
That is, using the usual notation for the field of fractions,
$$
K(x_1,x_2,\dots, x_n)=K(\{x_1,x_2,\dots, x_n\})
$$
and we are done.
