Hardness of FPT problems
Main idea of the answer: if we reduce an instance of parameterized Independent Set to parameterized Vertex Cover, then the parameter than we end up with depends on the size of the graph, and does not only depend on the input parameter. Now for some more detail.

As you know, a parameterized problem $Q$ is in (uniform) FPT if there is an algorithm that decides whether an input $(x,k)$ is contained in $Q$ in time $f(k) |x|^{O(1)}$ for some function $f$.

Since you can decide whether a graph $G$ has a vertex cover of size $k$ by picking an edge, and branching on which of its two endpoints to put in the vertex cover, this branching only goes $k$ deep (else you've put more than $k$ vertices in the cover), and easily runs in time $O(2^k n^2)$; therefore $k$-Vertex Cover is in FPT.

Now suppose we want to try use this algorithm to show that parameterized Independent Set is in FPT; assume we are given a graph $G$ on $n$ vertices and want to decide whether it has an independent set of size $\ell$. This is equivalent to asking whether $G$ has a vertex cover of size $n - \ell$. So we use our above algorithm to compute the answer in $O(2^{n - \ell} n^2)$ time. For our FPT algorithm, the exponential function in the running time may depend on the parameter, which is $\ell$, but it may NOT depend on the size of the input, which is $n$; but the approach we sketched uses time exponential in $n - \ell$ and is therefore not an FPT parameter with respect to the parameter $\ell$. This is why the fact that Vertex Cover is in FPT does not imply that Independent Set is in FPT.