The noether normalization is an inductive process on the number of generators of an affine $k$--algebra $A=k[x_1,\ldots,x_n]$.
At first you check if the $x_1,\ldots,x_n$ are algebraically independent over the field $k$. If they are, you are finished.
Otherwise you have a polynomial $f(x_1,\ldots,x_n) = 0$ that is fulfilled by the $x_i$.
Your next goal ist to find an automorphism $\phi:A \to A$ mit $x_i \mapsto y_i$, $y_i \in A$, so that $f(x_1,\ldots,x_n) = 0$ is mapped to $g(y_1,\ldots,y_n) = 0$ where $g$ has the form
$$(**) \quad\quad y_1^N + a_{1}(y_2,\ldots,y_n) y_1^{N-1} + \cdots + a_N(y_2,\ldots,y_n)$$
This shows you explicitly that $A=k[y_1,\ldots,y_n]$ is integral over the subring $B=k[y_2,\ldots,y_n]$ and you can now inductively proceed as $B$ is an affine $k$--algebra with one generator less.
Now back to the decisive step of choosing $\phi$, so that $y_1$ is "singled out" in $g(y_1,\ldots,y_n) = 0$.
There are essentially two substitutions used. One is more general and the other applies if $k$ is infinite (or algebraically closed).
Let $f(x_1,\ldots,x_n) = \sum a_{i_1,\ldots,i_n} x_1^{i_1} \cdot \cdots \cdot x_n^{i_n}$ and let without restriction of generality $x_1$ appear explicitly in $f$.
Then choose $x_1 = y_1$ and $x_i = y_i + y_1^{D^{i-1}}$. Then
$$
\begin{multline}
x_1^{i_1} \cdots x_n^{i_n} \mapsto y_1^{i_1} y_1^{D^1 i_2} y_1^{D^2 i_3}
\cdots y_n^{D^{n-1} i_n} + r(y_1,y_2,\ldots,y_n) = \\
y_1^{i_1 + D i_2 + D^2 i_3 + \cdots + D^{n-1} i_n} + r(y_1,\ldots,y_n)
\end{multline}
$$
where the power of $y_1$ in $r$ is lower than $N_{i_1,\ldots,i_n} = i_1 + D i_2 + \cdots + D^{n-1} i_n$.
Now if you choose $D$ bigger then the biggest occuring $i_\nu$ of the exponents, all the $N_{i_1,\ldots,i_n}$ are different (think $D$--adic integers) and the biggest one is the $N$ of the formula (**) above.
This was the most general substitutions, which works regardless of the nature of $k$.
The other one is $x_1 = y_1$ and $x_i = y_i + \alpha_i y_1$, where $\alpha_i \in k$. Then you get
$$
\begin{multline}
x_1^{i_1} \cdot \cdots \cdot x_n^{i_n} \mapsto \alpha_2^{i_2} \cdots \alpha_n^{i_n} y_1^{i_1+\cdots+i_n} + r(y_1,\ldots,y_n)
\end{multline}
$$
where the degree of $y_1$ in $r$ is less than $M_{i_1,\ldots,i_n} = i_1 + \cdots + i_n$. If now
$$f = f_d + f_{d-1} + \cdots + f_0$$
is the splitting into homogeneous components, then
$$g(y_1,\ldots,y_n) = f_d(1,\alpha_2,\ldots,\alpha_n) y_1^d + r(y_1,\ldots,y_n)$$
where the degree of $y_1$ in $r(y_1,\ldots,y_n)$ is less than $d$.
So you have again $g$ in the form $(**)$ when you choose $\alpha_i$ such that
$f_d(1,\alpha_2,\ldots,\alpha_n) \neq 0$ which is always possible if $k$ is infinite.
