This is an amplification of Mohan's suggestion to consider the context of projective varieties. I will asume that $k$ is infinite.

Given your $k$-algebra $R$, we get Spec $R$ as a variety $V$ in $\mathbb A^d$ for some $d$.
We may take the projective closure (i.e. Zariski closure) of $V$ in $\mathbb P^d$ to obtain a projective variety $\overline{V}$.

If $V = \mathbb A^d$ there is nothing to say: $R = k[x_1,\ldots,x_d]$ and witnesses its
own Noether normalization. So assume that $V$ is a proper subvariety of $\mathbb A^d$,
so that $\overline{V}$ is a proper subvariety of $\mathbb P^d$.

Because $k$ is infinite, we can (and do) choose a point $P$ (defined over $k$) lying at infinity, and not on $\overline{V}$. We can (and do) also choose a hyperplane $H$
(defined over $k$) which does not contain $P$. We may then
apply the projection from $P$ to $H$ to obtain a morphism $\pi: \overline{V}
\to H$. Properness of projective morphisms implies that this map has closed image
in $H$, say $\overline{W}$.

Because $P$ was chosen to lie at infinity, this restricts to a morphism
$V \to H \cap \mathbb A^d$, which again has closed image, say $W$. (As the notation
suggests, $\overline{W}$ will then be the projective closure of the affine variety
$W$.) This map is finite, in the sense that $V \to W$ corresponds to a map
$R' \to R$ with $R$ finite over $R'$.

Now if $W = H\cap \mathbb A^d$, we have obtained our Noether normalization of $R$,
since $R'$ is a polynomial ring in this case. If not, we proceed inductively,
replacing $V$ by $W$ and $\mathbb A^d$ by $H\cap \mathbb A^d$ (an affine space of
one dimension less). Eventually we will reach a stage where the projection
to the hyperplane is surjective. (If $V$ has dimension $n$ then we have to
perform $d -n$ projections altogether.)

So you see that you have a lot of flexibility in how to achieve the normalization
(because at each stage there are a lot of choices of $P$ and $H$), but the choices
are not completely arbitrary (because of the condition that $P$ not lie on
$\overline{V}$, and that $H$ not contain $P$). For example, thinking this way,
you will easily see what goes wrong in Mohan's counterexample. (Taking $a = x$
in his example corresponds to projecting from a point $P$ that *does* lies on
the projective closure of the hyperbola.)