Quotient of the ring of integers of a number field by a prime ideal
First, let us show that $\mathcal{O}_K$ is isomorphic to $\mathbb{Z}[x]/(f(x))$, with notation as in your question. Define a map: $$\psi:\mathbb{Z}[x]/(f(x)) \to \mathcal{O}_K$$ by sending $x$ to $\omega$, and extend by $\mathbb{Z}$-linearity, i.e., $$\psi(q(x) \bmod (f(x))) = q(\omega),$$ for any $q(x)\in\mathbb{Z}[x]$. Then, $\psi$ is well defined (if $q(x)$ and $q'(x)$ are congruent modulo $(f(x))$ then they differ by a multiple of $f(x)$, but $f(\omega)=0$), it is a ring homomorphism, and its kernel is trivial (if $q(\omega)=0$, then the minimal polynomial of $\omega$, which is $f(x)$, must divide $q(x)$). The map is clearly surjective ($\psi(a+bx)=a+b\omega$), so it is an isomorphism.
For your second question, notice that if $p$ is in the ideal in the quotient, then this affects the coefficients of every polynomial, and in effect reduces each coefficient modulo $p$. Thus, $$\mathbb{Z}[x]/(p,f(x)) \cong (\mathbb{Z}/p\mathbb{Z})[x]/(f(x) \bmod p\mathbb{Z}[x]) \cong \mathbb{F}_p[x]/(\tilde{f}(x)),$$ where $\tilde{f}(x)$ is the polynomial we obtain when we reduce each coefficient of $f(x)$ modulo $p$, and consider each coefficient in $\mathbb{F}_p$.
If $f(x)$ is irreducible over $\mathbb{F}_p[x]$, then $(\tilde{f}(x))$ is a prime ideal (and also a maximal ideal). Hence, the quotient $\mathbb{F}_p[x]/(\tilde{f}(x))$ is a field. Since $f(x)$ is of degree $2$, this field is isomorphic to $\mathbb{F}_{p^2}$. Hence, $$\mathcal{O}_K/(p) \cong \mathbb{Z}[x]/(p,f(x)) \cong \mathbb{F}_p[x]/(\tilde{f}(x)) \cong \mathbb{F}_{p^2}.$$