Pattern Recognition, Bishop - (Linear) Discriminant Functions 4.1
No, $y(\mathbf{x})=\mathbf{w}^T\mathbf{x}+w_0$; it is a scalar.  The dot product $\mathbf{w}^T\mathbf{x}$ is $\|\mathbf{w}\|$ times the length of the projection of $\mathbf{x}$ onto $\mathbf{w}$.  $w_0$ in your figure would be negative, and has the property that $y(\mathbf{x})=0$ whenever $\mathbf{x}$ is on the decision boundary.
No normalization appears to be necessary.  Certainly $r$ can be arbitrarily large (either positive or negative), when $\mathbf{x}$ is far away from the decision boundary.
As mentioned previously, the first term is actually a length from the origin; the bias serves to shift this so that $y$ itself is the orthogonal (scalar) component [up to scaling by $\|w\|$...But that's what we're out to show when we pass from Eq(4.6) to (4.7)].


The text's approach is to decompose $\mathbf{x}$ into components relative to the decision boundary: $\mathbf{x}_{\perp}$ on the boundary, and something perpendicular to the boundary.  Being perpendicular to the boundary, it is in the direction of $\mathbf{w}$, but we don't know how far, so they introduce its length as the unknown $r$.  (There's some standard geometry stuff here that could also get us to the conclusion, but I'll explain their approach.)

Now, as mentioned before, $y$ is zero on the boundary, so they have $y(\mathbf{x}_{\perp})=0$.  And now, just to fill in some of the details of what they say,
$$\begin{align*}
\mathbf{x} &amp;= \mathbf{x}_{\perp} + r \frac{\mathbf{w}}{\|\mathbf{w}\|}\\
\mathbf{w}^T\mathbf{x}+w_0 &amp;= \mathbf{w}^T\mathbf{x}_{\perp}+w_0 + r \frac{ \mathbf{w}^T\mathbf{w} }{ \|\mathbf{w}\| } \\
y(\mathbf{x}) &amp;= y(\mathbf{x}_{\perp}) + r \frac{\|\mathbf{w}\|^2}{\|\mathbf{w}\|} \\
y(\mathbf{x}) &amp;= 0 + r \|\mathbf{w}\|,
\end{align*}$$
and so $r=y(\mathbf{x})/\|\mathbf{w}\|$ as desired.

[I'm not sure what you meant in your last few lines; at least some of it seems to have been a typo?  Feel free to follow up.]



EDIT: Regarding your addition about substituting $r$, you should get $$\mathbf{x}=\mathbf{x}_{\perp}+y(\mathbf{x})\frac{\mathbf{w}}{\|\mathbf{w}\|^2},$$
but $\|\mathbf{w}\|^2$ is not equal to $\mathbf{w}$; the former is a scalar, and the latter a vector!  Rewriting, we have
$$\mathbf{x}=\mathbf{x}_{\perp}+\frac{y(\mathbf{x})}{\|\mathbf{w}\|}\frac{\mathbf{w}}{\|\mathbf{w}\|}.$$
This now looks correct: from the origin, go to $\mathbf{x}_{\perp}$, then along the unit vector $\mathbf{w}/\|\mathbf{w}\|$ for a distance of $y(\mathbf{x})/\|\mathbf{w}\|$ (which, per your bubble3, is the correct distance).