We'll localize at $\mathfrak p$ to make the base ring a PID. Let $A := \mathcal O_{K,\mathfrak p}$ be the localization of $\mathcal O_K$ at $\mathfrak p$, which really means localization at the complement $S = \mathcal O_K - \mathfrak p$. The ring $A$ is a PID with fraction field $K$ and one nonzero prime ideal $\mathfrak pA$, which is principal (generated by any element of $\mathfrak p - \mathfrak p^2$) and the natural map of residue fields $\mathcal O_K/\mathfrak p \to A/\mathfrak pA$ is an isomorphism.
Let $B$ be the integral closure of $A$ in $L$: it is the localization of $\mathcal O_L$ at $S$. The denominators of elements of $B$ are units in $A$, so the natural map $\mathcal O_L/\mathfrak p\mathcal O_L \to B/\mathfrak pB$ is an isomorphism.
Writing the reduction of $\alpha$ in $\mathcal O_L/\mathfrak p\mathcal O_L$ as $\overline{\alpha}$, the desired result $\mathcal O_L/\mathfrak p\mathcal O_L = (\mathcal O_K/\mathfrak p)[\overline{\alpha}]$ is the same as saying $B/\mathfrak pB = (A/\mathfrak pA)[\overline{\alpha}]$. The latter equality is what we'll prove, and in fact we're going to show $B = A[\alpha]$, which makes the equality of reductions obvious.
The hypothesis that $\mathfrak p \nmid ({\rm N}_{L/K}(f'(\alpha))\mathcal O_K$ implies $\mathfrak p$ and $({\rm N}_{L/K}(f'(\alpha))\mathcal O_K$ are relatively prime in $\mathcal O_K$ since $\mathfrak p$ is prime, so the extended ideals $\mathfrak pA$ and $({\rm N}_{L/K}(f'(\alpha))A$ are relatively prime in $A$. Thus, since $\mathfrak pA$ is the only nonzero prime ideal in $A$, the number ${\rm N}_{L/K}(f'(\alpha))$ is a unit in $A$. It is classical that ${\rm disc}(f) = {\rm N}_{L/K}(f'(\alpha))$, so
${\rm disc}(f)$ is a unit in $A$.
There is an important connection between ${\rm disc}(f)$ and the relation between $A[\alpha]$ and $B$. Recall the formula ${\rm disc}(\mathbf Z[\gamma]) = [\mathcal O_F:\mathbf Z[\gamma]]^2{\rm disc}(\mathcal O_F)$ where $F = \mathbf Q(\gamma)$ for an algebraic integer $\gamma$ and discriminants are computed using $\mathbf Z$-module bases of $\mathbf Z[\gamma]$ and $\mathcal O_F$. And when $\gamma$ has minimal polynomial $g(x)$ in $\mathbf Z[x]$ we have ${\rm disc}(\mathbf Z[\gamma]) = {\rm disc}(g)$, so
$$
{\rm disc}(g) = [\mathcal O_F:\mathbf Z[\gamma]]^2{\rm disc}(\mathcal O_F).
$$
This formalism works with $\mathbf Z$ replaced by $A$, $\mathbf Z[\gamma]$ replaced by $A[\alpha]$, and $\mathcal O_F$ replaced by $B$ as follows.
Let $n = \deg f$. Since $A$ is a PID and $L/K$ is separable, the integral closure $B$ of $A$ is a finite free $A$-module with rank $n$. Inside $B$ is the ring $A[\alpha]$, which is also a finite-free $A$-module with rank $n$.
Any ring $R$ containing $A$ that is a finite-free $A$-module has a discriminant defined by ${\rm disc}_A(R) = \det({\rm Tr}(e_ie_j))$ where
$\{e_1,\ldots,e_n\}$ is an $A$-basis of $R$. Changing that basis changes ${\rm disc}_A(R)$ by a unit square factor, so ${\rm disc}_A(R)$ is well-defined in general only up to unit squares. (In $\mathbf Z$ the only unit square is $1$, which is why discriminants over $\mathbf Z$ are independent of the choice of basis.) This is still enough to make the prime factors of ${\rm disc}_A(R)$ independent of the choice of basis.
We'll work with $A$-module discriminants of $A[\alpha]$ and $B$. They are related
$$
{\rm disc}_A(A[\alpha]) = [B:A[\alpha]]_A^2{\rm disc}_A(B),
$$
where $[B:A[\alpha]]_A$ is the $A$-module index of $A[\alpha]$ in $B$ (an element of $A$ defined up to unit multiple using the structure of the torsion $A$-module $B/A[\alpha]$ as a direct sum of cyclic $A$-modules). Since $f$ is the minimal polynomial of $\alpha$ over $A$, by using the obvious $A$-basis $\{1,\alpha,\ldots,\alpha^{n-1}\}$ of $A[\alpha]$ we get
$$
{\rm disc}_A(A[\alpha]) = {\rm disc}(1,\alpha,\ldots,\alpha^{n-1}) = {\rm disc}(f).
$$
Thus
$$
{\rm disc}(f) = [B:A[\alpha]]_A^2{\rm disc}_A(B).
$$
We already saw that $\mathfrak p \nmid {\rm N}_{L/K}(f'(\alpha))$ implies ${\rm disc}(f)$ is a unit in $A$. Thus $[B:A[\alpha]]_A$ is a unit in $A$, which means $B = A[\alpha]$.
