In Vakil's notes on Algebraic Geometry, he states in exercise 24.4.H that the normalisation of a singular curve is never flat and he claims that this is a simple consequence of the fact that every finite morphism $f:X\to Y$ is flat iff $f_*O_X$ is locally free.
Now I was thinking as follows. Denote the singular point of $Y$ by $y$. Then, $Y\setminus y\cong X\setminus f^{-1}(y)$, so $f$ is flat away from the point $y$, since isomorphisms are flat by definition. Thus consider $g=f\mid_{f^{-1}(y)}:f^{-1}(y)\to y$. $g$ is finite, thus to not be a flat map, we need to see that $g_*O_{f^{-1}(y)}$ is not locally free. Consider $$0\to O_y\to g_*O_{f^{-1}(y)}\to coker(g^*)\to 0.$$ If we are given $\phi\in O_y$, i.e. $\phi:U=U(y)\to k$, then $g^*\phi:g^{-1}(U)\to k$ gives an element of $g_*O_{f^{-1}(y)}$ and a map $g^{-1}(U)\to k$ is determined by its values on $g^{-1}(y)=f^{-1}(y)$ which consists of either one or two points, depending on whether we are given a node or not. The cokernel thus is a skyscraper sheaf supported at $y$, if we are dealing with a cusp, and trivial otherwise. In both situations, it seems that $g_*O_{f^{-1}(y)}$ is a free $O_y$-module of finite rank which would imply that $f$ is flat.
What am I doing wrong?