In the book Principles of Algebraic Geometry, there is a theorem:
Every meromorphic function on an algebraic variety $V\subset \mathbb{P}^n$ is rational.
And the proof of this assertion is in two stages: first, we express $V$ as a branched cover of a linear subspace $\mathbb{P}^k\subset \mathbb{P}^n$ by projection $\pi$, and deduce from this representation that the pullback $\pi^{*}K(\mathbb{P}^k)$ to $V$ of the field of rational functions on $\mathbb{P}^k$ has index at most $d= \operatorname{deg}(V)$ in the field of meromorphic functions $\mathfrak{M}(V)$; we then show that the field of rational functions $K(V)$ on $V$ is an extension of degree at least $d$ over $\pi^{*}K(\mathbb{P}^k)$.
I don't know why we can conclude the assertion? Since every meromorphic function $\varphi$ satisfies a polynomial relation of degree $d$ over fields of rational functions $\pi^{*}K(\mathbb{P}^k)$
$\varphi^d-\pi^{*}a_1 \varphi^{d-1}+\pi^{*}a_2 \varphi^{d-2}+...+(-1)^da_d \equiv0$
where $a_i$ are rational functions, we could say $\varphi$ is rational? Hope someone could help. Thanks!
The reason is, there are field extensions $$\pi^*K(\mathbb P^k)\subset K(V)\subset \mathcal{M}(V).$$
If one can show (1). $[\mathcal{M}(V):\pi^*K(\mathbb P^k)]\le d$ and (2). $[K(V):\pi^*K(\mathbb P^k)]\ge d$, then we must have $$[\mathcal{M}(V):K(V)]=1,$$ which implies $K(V)=\mathcal{M}(V),$ i.e., all meromorphic functions on $V$ are rational.
(1) follows from the identity $\phi^d-\pi^*a_1\phi^{d-1}+\cdots+(-1)^d\pi^*a_d\equiv 0, a_i\in K(\mathbb P^k)$ that you wrote, and (2) is standard in algebraic geometry.
P.S. There is a different approach to show $K(V)=\mathcal{M}(V)$ without using Noether normalization. See Theorem 8,3, 8.4, Shafarevich's Basic Algebraic Geometry, II.