could someone explain to me why the expected value of $y(n)$ is the following:
$\operatorname{E}(y(n)) = f(n)$
when $y(n) = x(n) + f(n)$ and $x(n)$ has zero mean.
But why is the expected value of a deterministic sequence ($f(n)$) the sequence itself?
The expectation operator is always taken over a random variable. In your case, the random variable is $x(n)$.
Now, $$E\left[f(n)\right] = \int\limits_{-\infty}^{\infty} f(n) p_X(x) dx = f(n)\int\limits_{-\infty}^{\infty} p_X(x) dx = f(n) .$$ You can take the deterministic number out of the integral since it does not depend on the variable of integration. The last equality is because the area under a distribution is unity.