Let $k$ be a field and $$\varphi: A:= k[t] \to B$$ a finite injective ring morphism of integral domains. Obviously $\operatorname{Frac}(A)=k(t)$. Assume that $\operatorname{Frac}(B)= k(t)[y]=k(t)[Y]/(Y^2-P(t))$ with $P(t) \in k[t]$ non constant and without a square factor.
I wish to show that $B=k[t,y]=k[t,Y]/(Y^2-P(t))$ holds and still struggling with verification of the inclusion $$k[t,Y]/(Y^2-P(t)) \subset B.$$
The inclusion $B \subset k[t,Y]/(Y^2-P(t)) $ I have already proved:
I'm using as black box that $k[t,y]=k[t,Y]/(Y^2-P(t))$ is normal. the proof is really awkward and I would like to skip it at this point. therefore by definition $k[t,y]=k[t,Y]/(Y^2-P(t))$ is nothing but integral closure of $\operatorname{Frac}(B)=k(t)[Y]/(Y^2-P(t))$. this also implies that $k[t,y]$ is the integral closure of $A=k[t]$ in $\operatorname{Frac}(B)$.
by assumption $\phi:k[t] \to B$ is is finite and therefore integral. this implies that $B \subset k[t,y]$ by UP of normalization.
the "$k[t,Y]/(Y^2-P(t)) \subset B$" direction I haven't still succeed. since $\operatorname{Frac}(B) \neq \operatorname{Frac}(A)$ we have $k[t] \cong \varphi(A) \neq B$ therefore $B$ contains an element of the shape $y \cdot F(t) \in k[t,y]$ with certain $F(t) \in k[t]$.
how to verify that $y \in B$?
This doesn't have to hold if $B$ is not integrally closed. For example, if $B=k[t,a]/(a^2-t^3)$, your hypotheses hold: $\operatorname{Frac}(B)$ is generated over $k[t]$ by $Y=a/t$, which has minimal polynomial $x^2-t$, so $\operatorname{Frac}(B)=k(t)[Y]/(Y^2-t)$.
Now, you can't prove that $B\subset k(t)[Y]/(Y^2-t)$ because $Y$ is not contained in $B$. This can be seen because $B=k[t,a]=k[Y^2,Y^3]\subsetneq k[Y]=k[t,Y]/(Y^2-t)$.