Why isn't $g^{mu nu}partial_{nu}$ equal to $partial^{mu}$ in General Relativity?

Well I'll answer but I would be interested to know if my answer is wrong. I thought we defined $partial^a$ such that $$ partial^a f equiv g^{amu} partial_mu f . $$ The thing to be careful about is then the fact that $$ partial_mu (g^{amu} f) = (partial_mu g^{amu}) f + g^{amu} partial_mu f $$ so usually $$ partial_mu (g^{amu} f) ne partial^a f . $$ It also follows that you cannot assume that $partial^a$ operators commute, nor even that $partial^mu partial_mu$ is equal to $partial_mu partial^mu$ unless you make some other definition of $partial^a$.

For a scalar function; $nabla_{mu}f=partial_{mu}f$, its a definition, we know $nabla_{mu}f$ is a $(0,1)$ tensor so $g^{munu}nabla_{mu}f=nabla^{nu}f$ but there doesn't exist any similar definition where we identify contravariant version of $nabla^{mu}f$ with $partial^{mu}f$. So ultimately it boils down to the definition of $partial^{mu}$.

The definition $$ partial^{mu} = g^{mu nu} , partial_{nu} tag{1} $$ doesn't have more sense than $$ x_{mu} = g_{mu nu} , x^{nu}. tag{2} $$ Using (1) may lead to many troubles that you want to avoid at all cost. Don't use it!

However, you may use the same for the covariant derivative: $$ nabla^{mu} = g^{mu nu} , nabla_{nu}, tag{3} $$ which may be usefull since $nabla_{lambda} , g^{mu nu} = 0$.