Let $L/k$ be any field extension and suppose $K/k$ is finitely generated and of transcendence degree $1$, i.e. we find $T\in K$ transcendental over $k$ such that $K$ is a finite extension of $k(T)$. I am trying to prove the following: $L\otimes_k K$ is a finite dimensional $L(X)$-algebra (where $L(X)$ is the fraction field of the polynomial ring $L[X]$).
I am having problem even defining the structure map $L(X)\rightarrow L\otimes_k K$. My idea was that the inclusion $L\rightarrow L\otimes_k K$ induces an inclusion $L[X]\rightarrow L\otimes_k K$ by sending $X\mapsto 1\otimes T$, then if every nonzero $f\in L[X]$ becomes invertible in $L\otimes_k K$ then the universal property of localization yields the desired map $L(X)\rightarrow L\otimes_k K$, but I don't know how to prove this. Certainly $1\otimes T$ is invertible, but this is not sufficient.
Furthermore, even if what I just wrote makes sense, I am not sure how to proceed regarding finiteness. Certainly $K\cong k(T)^n$ as a $k(T)$-vector space, hence $L\otimes_k K\cong (L\otimes_k k(T))^n$ as a $L\otimes_k k(T)$-vector space, but how prove that the latter is a finite-dimensional $L(X)$-vector space? This boils down to the same problem I had with defining the $L(X)$-algebra structure in the first place.
It's generally not true that $L \otimes_k K$ is a $L(X)$-algebra, even with additional separability assumptions. (I'm assuming that you want the $L(X)$-algebra to extend the obvious $L$-algebra structure here)
Suppose $K=k(T)$, then we get with $S=k[T] \setminus \{0\}$ $$L \otimes_k K \cong L \otimes_k k[T] \otimes_{k[T]} k(T) \cong L[T] \otimes_{k[T]} k(T) = L[T] \otimes_{k[T]} S^{-1}k[T] = S^{-1}L[T]$$
This equals $L(T)$ iff $L/k$ is algebraic (then the extension $L[T] / k[T]$ is integral).
Generally if $K/k$ is any extension and $T \in K$ is transcendental, then we have an inclusion $k(T) \subset K$ which gives an injection $L \otimes_k k(T) \hookrightarrow L \otimes_k K$ since every module over a field is flat. If $K / k(T)$ is an algebraic extension (which is equivalent to $K$ having transcendence degree $1$), then $L \otimes_k K$ is an integral extension of $L \otimes_k k(T)$. Integral ring extensions $R \hookrightarrow A$ have the property that if $r \in R$ is invertible in $A$, then it is already invertible in $R$, because in this situation, extensions of proper ideals are proper. Thus the question whether elements of $L[T]$ are invertible in $L \otimes_k K$ reduces to the question whether elements of $L[T]$ are invertible in $L \otimes_k k(T)$ which is answered above.
To summarize: $L \otimes_k K$ is a $L(X)$ algebra for any transcendental $X$ iff $L/k$ is algebraic.
If we suppose that $L \otimes_k K$ is indeed a $L(X)$-algebra, then we must already have $L \otimes_k k(T) = L(T)$, so finiteness follows from your observation that $L \otimes_k K$ is a finitely generated $L \otimes_k k(T)$-module if we take $X=T$.