It's worth noting that the cone, itself, cannot be written in the form $$\vec r(t)=t\hat i+2t\cos(t)\hat j+2t\sin(t)\hat k.$$
The question at hand is, does that curve lie on the cone? In particular, all we need to show is that for $\langle x,y,z\rangle$ on that curve--that is, $x=t,y=2t\cos(t),z=2t\sin(t)$ for some $t$--we have $y^2+z^2=4x^2$, and the book wrote out precisely enough steps to show that.
You actually had just about everything right, except that you skipped an important step: your normal vector to the surface $ \ \vec{ds} \ $ is correct, but you need to integrate its length over the surface of the cone nappe in order to obtain the surface area.
I'll generalize the problem a little, since the choice of proportions for the cone hides one of the factors in the surface area result. For a cone nappe with a height $ \ h \ $ and a "base radius" $ \ r \ $ , we can use similar triangles to find the parametrization (using your notation)
$$ x \ = \ \left( \frac{r}{h} \right) u \ \cos \ p \ \ , \ \ y \ = \ \left( \frac{r}{h} \right) u \ \sin \ p \ \ , \ \ z \ = \ u \ \ , $$
with the domain $ \ 0 \ \le \ u \ \le \ h \ , \ 0 \ \le \ p \ < \ 2 \pi \ $ . An "upward" normal vector is then given by
$$ \vec{R_u} \ \times \ \vec{R_p} \ \ " = " \ \ \left|\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ \left( \frac{r}{h} \right) \cos \ p&\left( \frac{r}{h} \right) \sin \ p\quad&1\\ -\left( \frac{r}{h} \right) u \ \sin \ p&\left( \frac{r}{h} \right) u \ \cos \ p\quad&0\end{array}\right| $$
$$ = \ \langle \ -\left( \frac{r}{h} \right) u \ \cos \ p \ \ , \ \ -\left( \frac{r}{h} \right) u \ \sin \ p \ \ , \ \ \left( \frac{r}{h} \right)^2 u \ \rangle \ \ . $$
So, up to this point, your procedure is fine. What is needed now is the "norm" of this vector:
$$ \| \ \vec{R_u} \ \times \ \vec{R_p} \ \| \ \ = \ \ \left[ \ \left( \frac{r}{h} \right)^2 u^2 \ \cos^2 \ p \ + \ \left( \frac{r}{h} \right)^2 u^2 \ \sin^2 \ p \ + \ \left( \frac{r}{h} \right)^4 u^2 \ \right]^{1/2} \ \ . $$
$$ = \ \ \left[ \ \left( \frac{r}{h} \right)^2 u^2 \ + \ \left( \frac{r}{h} \right)^4 u^2 \ \right]^{1/2} \ = \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \ u \ \ . $$
It is the "magnitude" of the infinitesimal patches associated with the normal vectors that we wish to integrate over the domain of the parameters. Thus,
$$ S \ \ = \ \ \int_0^{2 \pi} \int_0^h \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \ u \ \ du \ dp $$
$$ = \ \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \int_0^{2 \pi} dp \ \int_0^h \ u \ \ du $$
$$ = \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \cdot \ 2 \pi \ \cdot \ \left(\frac{1}{2}u^2 \right) \vert_0^h \ \ = \ \left(\frac{r}{h} \right) \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ \cdot \ \pi \ h^2 $$
$$ = \ \pi \ h \ \cdot \ \left(\frac{r}{h} \right) \ \cdot \ h \ \sqrt{ 1 \ + \ \left( \frac{r}{h} \right)^2 } \ = \ \pi \ r \ \sqrt{ r^2 \ + \ h^2 } \ \ , $$
or $ \ \pi \ $ times the "base radius" times the "slant height" of the cone nappe, as the surface area is frequently expressed. In your use of the "standard cone", for which $ \ r \ = \ h \ $ , this formula gives us $ \ S \ = \ \pi \ \sqrt{2} \ h^2 \ $ , as you will find for your calculations, with the restoration of the omitted step.
Best Answer
One easy way to solve this is to use Cartesian coordinates to get the relations between the variables, then to translate that to a vector function.
In this case, you have two expressions for $z$. Equate them and we get
$$\sqrt{x^2+y^2}=1+y$$
$$x^2+y^2=1+2y+y^2$$
$$y=\frac 12x^2-\frac 12$$
We need to be careful of not introducing extraneous solutions in our squaring the equation, but none seem to be in evidence here.
We can set $x$ to the parameter and use the second of your equations to get
$$\vec{r}(t)=(t)\vec{i} + \left(\frac 12t^2-\frac 12\right)\vec{j} + \left(\frac 12t^2+\frac 12\right)\vec{k}$$
There are, of course, infinitely many other possibilities.
As a quick geometric check, we see that the problem asks for the intersection of a cone with a plane parallel to an "edge" of the cone. The ancient Greeks knew that was nothing or a parabola. The point $(1,0,1)$ is in the intersection, so the correct answer here is not nothing. Our answer gives a "tilted parabola" so it fits.