Is this a Fixed Parameter Tractable algorithm? Suppose my algorithm runs in time $O(nL^2)$, where $n$ is the size of the input, and $L$ some other parameter, which can get arbitrarily large w.r.t. $n$. My algorithm does not run in polynomial time, since $L$ can get arbitrarily large w.r.t. $n$. So I'd like to categorize it elsewise.
Fixed Parameter Tractable algorithms, with regards to some defined parameter $k$, run in time $O(f(k)\times n^{O(1)})$, where $f$ is any function. 

Would it be correct to state that my algorithm is Fixed Parameter Tractable (FPT) with regards to the parameter $L$?
First and foremost, I would advise having another look at the basic definitions. An FPT instance is identified by a string $x$ and an integer $k$ such that $(x, k)$ is a yes-instance. 

Put briefly, if $f(k)$ is not polynomial in terms of $n$ and still depends on $n$, then you cannot obtain an FPT algorithm. We require that $k$ is fixed, or independent of $n$, and thus $f(k)$ is always independent of $n$. However, if $f(k) = n^{O(1)}$, then our FPT algorithm is in fact a polynomial time algorithm. Hence, if $f(k) = O(2^n)$, the answer is definitely no.