I think I know how to answer this question now! Our plan is first to show $K(x)[y] / $ is a field and then to show $K(x)[y] / \cong$ Frac$(K[x, y] / )$ by using the universal property of field of fractions. From now on, I will denote the ideal generated by $f$ using $$ since I will be using a lot of parenthesis.
To show $K(x)[y] / $ is a field, it suffices to show $f \in K(x)[y]$ is irreducible. We will use the generalized Eisenstein's criterion. Note $x - \alpha_j \in K[x]$ being irreducible $\implies $ is a prime ideal in $K[x]$. Now $-\prod_{i = 1}^{d - 1}(x - \alpha_i) \in $, $0 \in $, $1 \notin $ and $-\prod_{i = 1}^{d - 1}(x - \alpha_i) \notin ^2 = <(x - \alpha_j)^2>$. Here we assume $\alpha_i$ are distinct, which I guess is an unstated assumption. So by the generalized Eisenstein's criterion, $f \in K[x][y]$ is irreducible. Since $K[x]$ is a UFD, $f \in K(x)[y]$ is also irreducible. Hence $K(x)[y] / $ is a field.
We are left to show that $K(x)[y] / $ is the smallest field containing $K[x, y] / $ up to isomorphism, i.e. it posses the universal property that for any embedding $\sigma: K[x, y] / \hookrightarrow \Omega$, where $\Omega$ is a field, we can find some embeddings $\iota: K[x, y] / \hookrightarrow K(x)[y] / $ and $j:K(x)[y] / \hookrightarrow \Omega$ such that $\sigma = j \circ \iota$. Before proceeding, we first observe $K[x, y] \cong K[x][y]$, so we can replace all instances of $K[x, y]$ with $K[x][y]$. Also, we will write the coset $a + I$ as $\overline{a}$.
Given embedding $\sigma: K[x][y] / \hookrightarrow \Omega$ by $$\overline{g_0(x) + g_1(x) y + \dots + g_{n - 1}(x) y^{n - 1}} \mapsto \sigma(\overline{g_0(x) + g_1(x) y + \dots + g_{n - 1}(x) y^{n - 1}})$$
with $g_i(x) \in K[x]$.
We define $\iota: K[x][y] / \hookrightarrow K(x)[y] / $ by $$\overline{g_0(x) + g_1(x) y + \dots + g_{n - 1}(x) y^{n - 1}} \mapsto \overline{\frac{g_0(x)}{1} + \frac{g_1(x)}{1} y + \dots + \frac{g_{n - 1}(x)}{1} y^{n - 1}}$$ with $g_i(x) \in K[x]$
and $j: K(x)[y] / \hookrightarrow \Omega$ by $$\overline{\frac{g_0(x)}{h_0(x)} + \frac{g_1(x)}{h_1(x)} y + \dots + \frac{g_{n - 1}(x)}{h_{n - 1}(x)} y^{n - 1}} \mapsto \frac{\sigma(\overline{g_0(x)})}{\sigma(\overline{h_0(x)})} + \frac{\sigma(\overline{g_1(x))}}{\sigma(\overline{h_1(x)})} \sigma(\overline{y}) + \dots + \frac{\sigma(\overline{g_{n - 1}(x)})}{\sigma(\overline{h_{n - 1}(x)})} \sigma(\overline{y})^{n - 1}$$ with $g_i(x), h_i(x) \in K[x]$.
Of course, it is then an exercise to check that $\iota, j$ are well-defined embeddings such that $j \circ \iota = \sigma$ :) Finally, since $K(x)[y] / $ satisfies the universal property of Frac$(K[x][y] / )$ and universal objects are isomorphic, we have $K(x)[y] / \cong$ Frac$(K[x][y] / ) \cong$ Frac$(K[x, y] / )$.
EDIT-1: We will check $j$ is injective. Suppose $$\frac{\sigma(\overline{g_0(x)})}{\sigma(\overline{h_0(x)})} + \frac{\sigma(\overline{g_1(x))}}{\sigma(\overline{h_1(x)})} \sigma(\overline{y}) + \dots + \frac{\sigma(\overline{g_{n - 1}(x)})}{\sigma(\overline{h_{n - 1}(x)})} \sigma(\overline{y})^{n - 1} = 0$$ From now on we write $g_i(x), h_i(x)$ as $g_i, h_i$. Then $$\frac{\sigma(\overline{g_0 h_1 \dots h_{n - 1} + h_0 g_1 h_2 \dots h_{n - 1} y + \dots + h_0 \dots h_{n - 2} g_{n - 1} y^{n - 1}})}{\sigma(\overline{h_0 \dots h_n})} = 0$$ We have $$\sigma(\overline{g_0 h_1 \dots h_{n - 1} + h_0 g_1 h_2 \dots h_{n - 1} y + \dots + h_0 \dots h_{n - 2} g_{n - 1} y^{n - 1}}) = 0$$ Since $\sigma$ is injective, $$\overline{g_0 h_1 \dots h_{n - 1} + h_0 g_1 h_2 \dots h_{n - 1} y + \dots + h_0 \dots h_{n - 2} g_{n - 1} y^{n - 1}} = 0$$ Since $\overline{h_i}$ are nonzero and $K(x)[y] / $ is a field (it is what we've just proved!), $\overline{h_1 \dots h_{n - 1}}$ is nonzero and hence a unit in $K(x)[y] / $. Thus we can divide the whole thing by $\overline{h_1 \dots h_{n - 1}}$ and obtain $$\overline{\frac{g_0}{h_0} + \frac{g_1}{h_1} y + \dots + \frac{g_{n - 1}}{h_{n - 1}} y^{n - 1}} = 0$$
