Since Matt proved (1) and (2), I'll prove the rest.
(3)
Let $RI^+(A)$ be the set of regular ideals of $A$.
Clearly $RI^+(A)$ is an ordered commutative monoid with mulitiplications of ideals.
Let $RI^+(B)$ be the set of ideals of $B$ which are relatively prime to $\mathfrak{f}$.
$RI^+(B)$ is also an ordered commutative monoid.
By (1) and (2), $RI^+(A)$ is canonically isomorphic to $RI^+(B)$ as an ordered commutative monoid.
Since $B$ is a Dedekind domain, (3) follows immediately.
(4) follows immediately from (3) and the following lemma.
Lemma 1
Let $P$ be a maximal ideal of $A$.
$P$ is invertible if and only if $P$ is regular.
Proof:
Suppose P is regular.
By this, $A_P$ is integrally closed.
Since $A_P$ is integrally closed, Noetherian and of dimension 1, it is a discrete valuation ring.
Hence $PA_P$ is principal.
Let $Q$ be a maximal ideal such that $Q \neq P$.
Since $P$ is not contained in $Q$, $PA_Q = A_Q$.
Hence $PA_Q$ is also principal.
Since $A$ is Noetherian, $P$ is finitely generated over $A$.
Hence P is invertible by this.
Suppose conversely P is invertible.
By this, $PA_P$ is principal.
Hence $A_P$ is a discrete valuation ring(e.g Atiyah-MacDonald).
Hence $A_P$ is integrally closed.
By this, $P$ is regular.
QED
Lemma 2
Let $A$ be a commutative Noetherian ring.
Let $I$ be a proper ideal of $A$ such that $dim A/I = 0$.
Then $A/I$ is canonically isomorphic to $\prod_P A_P/IA_P$, where $P$ runs over all the maximal ideals of $A$ such that $I \subset P$.
This is well knowm.
Lemma 3
Let $A$ be a Noetherian domain of dimension 1.
Let $I$ be a non-zero proper ideal of $A$.
Then $(A/I)^*$ is canonically isomorphic to $\bigoplus_{\mathfrak{p}} (A_\mathfrak{p}/IA_\mathfrak{p})^*$ as abelian groups, where $\mathfrak{p}$ runs over all the maximal ideals of $A$ such that $I \subset \mathfrak{p}$.
This follows immediately from Lemma 2.
Lemma 4
Let $A, K, B, \mathfrak{f}$ be as in the title question.
Then $(B/\mathfrak{f})^*$ is canonically isomorphic to $\bigoplus_{\mathfrak{p}} (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*$ as abelian groups, where $\mathfrak{p}$ runs over all the maximal ideal of $A$.
Proof:
By Lemma 3, $(B/\mathfrak{f})^*$ is canonically isomorphic to $\bigoplus_{\mathfrak{P}} (B_\mathfrak{P}/\mathfrak{f}B_\mathfrak{P})^*$ as an abelian group, where $\mathfrak{P}$ runs over all the maximal ideal of $B$ such that $\mathfrak{f} \subset \mathfrak{P}$.
If $\mathfrak{p}$ is a regular prime ideal of $A$, $A_\mathfrak{p}$ is integrally closed by this.Hence $B_\mathfrak{p} = A_\mathfrak{p}$.
Since $\mathfrak{f}A_\mathfrak{p} = A_\mathfrak{p}$, $B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p} = A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p} = 0$.
Hence we only need to consider $\mathfrak{p}$ such that $\mathfrak{f} \subset \mathfrak{p}$.
It's easy to see that $B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p}$ is canonically isomorphic to $\prod B_\mathfrak{P}/\mathfrak{f}B_\mathfrak{P}$, where $\mathfrak{P}$ runs over all the maximal ideals of $B$ lying over $\mathfrak{p}$.
Hence $(B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*$ is canonically isomorphic to $(\bigoplus B_\mathfrak{P}/\mathfrak{f}B_\mathfrak{P})^*$, where $\mathfrak{P}$ runs over all the maximal ideals of $B$ lying over $\mathfrak{p}$. QED
Lemma 5
Let $B$ be an integral domain.
Let $A$ be a subring of $B$ such that $B$ is integral over $A$.
Let $I$ be an ideal of $A$.
Let $\mathfrak{p}$ be a prime ideal of $A$ such that $I \subset \mathfrak{p}$.
Let $B_\mathfrak{p}$ be the localization of $B$ with respect the multiplicative subset $A - \mathfrak{p}$. Let $f:B_\mathfrak{p} \rightarrow B_\mathfrak{p}/IB_\mathfrak{p}$ be the canonical homomorphism.
$f$ induces a group homomorphism $g: (B_\mathfrak{p})^* \rightarrow (B_\mathfrak{p}/IB_\mathfrak{p})^*$.
Then $g$ is surjective.
Proof:
Since $A_\mathfrak{p}$ is a local ring and $B_\mathfrak{p}$ is integtral over $A_\mathfrak{p}$,
every maximal ideal $\mathfrak{Q}$ of $B_\mathfrak{p}$ lies over $\mathfrak{p}A_\mathfrak{p}$.
Hence $\mathfrak{Q}$ = $\mathfrak{P}B_\mathfrak{p}$, where $\mathfrak{P}$ is a maximal ideal of $B$ lying over $\mathfrak{p}$.
Since $I \subset \mathfrak{p}$, $I \subset \mathfrak{P}$.
Hence $IB_\mathfrak{p} \subset \mathfrak{Q}$.
Let $x \in B_\mathfrak{p}$.
Suppose $f(x)$ is invertible.
Then $f(x)$ is not contained in any maximal ideal of $B_\mathfrak{p}/IB_\mathfrak{p}$.
Suppose $x$ is not invertible.
$x$ is contained in a maximal ideal of $B_\mathfrak{p}$.
This is a contradiction. QED
Lemma 6
Let $A, K, B, \mathfrak{f}$ be as in the title question.
Let $\mathfrak{p}$ be a prime ideal of $A$ such that $\mathfrak{f} \subset \mathfrak{p}$.
Since $\mathfrak{f}$ is an ideal of both $A$ and $B$, $\mathfrak{f} = \mathfrak{f}A = \mathfrak{f}B$.
Hence $\mathfrak{f}A_\mathfrak{p} = \mathfrak{f}B_\mathfrak{p}$.
Hence $A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p} \subset B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p}$.
Hence $(A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^* \subset (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*$.
We claim $(B_\mathfrak{p})^*/(A_\mathfrak{p})^*$ is isomorphic to $(B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*/(A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^*$.
Proof:
By lemma 5, $g: (B_\mathfrak{p})^* \rightarrow (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*$
is surjective.
Let $\pi: (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^* \rightarrow (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*/(A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^*$ be the canonical homomorphism.
Let $h: (B_\mathfrak{p})^* \rightarrow (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*/(A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^*$ be $\pi g$.
Let $x \in (B_\mathfrak{p})^*$.
Suppose $h(x) = 0$.
Then $g(x) \in (A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^*$.
Hence thers exists $y \in A_\mathfrak{p}$ such that $x \equiv y$ (mod $\mathfrak{f}B_\mathfrak{p}$).
Since $\mathfrak{f}B_\mathfrak{p} = \mathfrak{f}A_\mathfrak{p}$, $x \in A_\mathfrak{p}$.
Since $g(x) \in (A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^*$, $x \in (A_\mathfrak{p})^*$.
Hence Ker$(h) = (A_\mathfrak{p})^*$.
QED
(6)
Let $A, K, B, \mathfrak{f}$ be as in the title question.
There exists the following exact sequence of abelian groups.
$0 \rightarrow B^*/A^* \rightarrow (B/\mathfrak{f})^*/(A/\mathfrak{f})^* \rightarrow I(A)/P(A) \rightarrow I(B)/P(B) \rightarrow 0$
Proof:
By this, there exists the following exact sequence of abelian groups.
$0 \rightarrow B^*/A^* \rightarrow \bigoplus_{\mathfrak{p}} (B_{\mathfrak{p}})^*/(A_{\mathfrak{p}})^* \rightarrow I(A)/P(A) \rightarrow I(B)/P(B) \rightarrow 0$
Here, $\mathfrak{p}$ runs over all the maximal ideals of $A$.
If $\mathfrak{p}$ is a regular prime ideal of $A$, $A_\mathfrak{p}$ is integrally closed.
Hence $B_\mathfrak{p} = A_\mathfrak{p}$.
Hence, in $\bigoplus_{\mathfrak{p}} (B_{\mathfrak{p}})^*/(A_{\mathfrak{p}})^*$,
it suffices to consider only $\mathfrak{p}$ such that $\mathfrak{f} \subset \mathfrak{p}$.
By Lemma 3,
$(A/\mathfrak{f})^*$ is canonically isomorphic to $\bigoplus_\mathfrak{p} (A_\mathfrak{p}/\mathfrak{f}A_\mathfrak{p})^*$ as an abelian group, where $\mathfrak{p}$ runs over all the maximal ideals of $A$ such that $\mathfrak{f} \subset \mathfrak{p}$.
By Lemma 4,
$(B/\mathfrak{f})^*$ is canonically isomorphic to $\bigoplus_{\mathfrak{p}} (B_\mathfrak{p}/\mathfrak{f}B_\mathfrak{p})^*$ as an abelian group, where $\mathfrak{p}$ runs over all the maximal ideal of $A$
such that $\mathfrak{f} \subset \mathfrak{p}$.
Now, by Lemma 6, we are done. QED
Lemma 7
Let $A, K, B, \mathfrak{f}$ be as in the title question.
Let $\phi: I(A)/P(A) \rightarrow I(B)/P(B)$ be the canonical homomorphism.
Let $C \in$ Ker($\phi$).
Then $C$ contains an ideal of the form $A \cap \beta B$, where $\beta$ is an element of $B$ such that $\beta B + \mathfrak{f} = B$.
This follows immediately from (6).
(5)
Let $I$ be an invertible ideal of $A$.
Since $B$ is a Dedekind domain, by this, there exist an ideal $\mathfrak{J}$ of $B$ and $\gamma \in K$ such that $\mathfrak{J} + \mathfrak{f} = B$
and $IB = \mathfrak{J}\gamma$.
Let $J = A \cap \mathfrak{J}$.
By (2), $J$ is regular and $JB = \mathfrak{J}$.
By (4), $J$ is invertible.
Since $IB = \mathfrak{J}\gamma = J\gamma B$, $IJ^{-1}B = \gamma B$.
By Lemma 7, there exists $\beta \in B$ such that $\beta B + \mathfrak{f} = B$ and $IJ^{-1} \equiv A \cap \beta B$ mod($P(A)$).
Hence $I \equiv J(A \cap \beta B)$ mod($P(A)$).
Since $J$ and $A \cap \beta B$ are regular, we are done.