How calculate with Riemannian metrics (e.g. Multiplication and Divison)? I have no idea how to handle the following Riemannian metrics, how to find the estimates for the bound and how to actually calculate with $g$ and $d$. Do I need to use the matrix representation? Or how do I multiply, divide etc with line elements metrics?
Here is what I have defined and calculated so far:
The finite paraboloid with height $0\leq h<\infty$ is defined as the set $M_{h}:=\{(x_{1},\, x_{2},\, x_{3}):\, h\cdot(x_{1}^{2}+x_{2}^{2})=x_{3},\,0<x_{3}\leq h\}$.
If we write the parameneter equations:
$x_{1}=\rho cos(\varphi)$
$x_{2}=\rho sin(\varphi)$
$x_{3}=\rho^{2},$
where $\rho\in[0,\sqrt{h}]$ and $\varphi\in]0,2\pi]$, then we find the induced metric $g$ on $M$ obtained by pulling back the flat metric $ds^{2}=dx_{1}^{2}+dx_{2}^{2}+dx_{3}^{2}$ on $\mathbb{R}^{3}$:
$g=(d\rho\cdot cos(\varphi))^{2}+(d\rho\cdot sin(\varphi))^{2}+(d\rho^{2})^{2}=(1+4\rho^{2})d\rho^{2}+\rho^{2}d\varphi^{2}$.
The unit disk $N=\{(y_{1},\, y_{2}):\, y_{1}^{2}+y_{2}^{2}\leq1\}$ is a 2-dimensional submanifold with boundary in $\mathbb{R}^{3}$. 
We define the parameter equations on $N$ by:
$y_{1}=R\cdot cos(\phi)$
$y_{2}=R\cdot sin(\phi)$,
where $R\in[0,1]$ and $\phi\in]0,2\pi]$.
Then the induced metric on $N$ is given by $d=dR^{2}+R^{2}d\phi^{2}$.
I then defined the diffeomorphism $F$ as follows as a projection :
$F:\, M\rightarrow N$
$(\rho,\,\varphi)\longmapsto(\frac{1}{\sqrt{h}}\rho,\,\varphi)=(R,\phi)$.
Then I pulled back the metric $d$ on $N$ and get the pulled-back metric $d*$ on $M$:
$d*=\frac{1}{\sqrt{h}}d\rho^{2}+1d\varphi^{2}$.
Now I am ready to show that $M$ and $N$ are quasi-isometric with certain constants $C,D$, but I am completely stuck here.
My goal is to show the following:
$\exists C,D$ such that $\forall x\in M$, $\forall v\in T_{x}M (v\neq0)$:
$C\leq\frac{g_{1_{h}}(x)(v,v)}{g_{2_{h}}(x)(v,v)}\leq D$ (- and if possible $C,D\approx1$).
 A: You question doesn't make full sense to me: a metric is a quadratic form, yet your $d$ has some linear terms in it. Also it's not clear exactly what you want to bound since at the top of your post you have two arbitrary vectors and at the bottom you have $g_1(v,v)$.
But let's assume that you want to bound $g_1(v,v)/g_2(v,v)$ over all vectors $v$ and points on your manifold. Let's first look at a fixed point on the manifold. The following fact will be useful:
If $A,B$ are positive-definite matrices, then
$$f(x) = \frac{x^TAx}{x^TBx} \in [\lambda_1, \lambda_n]$$
where $\lambda_1$ and $\lambda_n$ are the smallest and largest eigenvalue of $B^{-1}A$, respectively; and moreover these bounds are tight.
To see that the bounds are tight notice that $f(v_i) = \lambda_i$ for any eigenvector $v_i$ of $B^{-1}A$. Finally notice that when $v = \sum \alpha_i v_i$,
\begin{align*}
Av &= \sum \alpha_i \lambda_i Bv_i\\
&\leq \lambda_N Bv\\
v^TAv &\leq \lambda_N v^TBv
\end{align*}
and similarly $v^TAv \geq \lambda_1 v^TBv.$
Now use this fact to compute the bounds on the quotient at any point of your manifold, then bound the bounds as the point varies over the manifold.
A: For this particular calculation, it's perhaps easier to use Cartesian coordinates than to use polar coordinates (which degenerate at the origin), but it's even easier to use elementary calculus and geometry:
Let $(M, d')$ be the unit disk with the Euclidean metric, and let $(M, d)$ be the unit disk with the metric induced by the embedding
$$
(x, y) \mapsto (x, y, x^{2} + y^{2}).
$$
This map is a graph parametrization, so obviously
$$
d'(p, q) \leq d(p, q)\quad\text{for all $p$, $q$ in $M$.}
$$
To get an upper bound, note that the gradient of the quadratic function has magnitude at most $2$ in the unit disk, so the Pythagorean theorem gives
$$
d(p, q) \leq \sqrt{5}\, d'(p, q)\quad\text{for all $p$, $q$ in $M$.}
$$
Analytically, the Riemannian metric on the paraboloid (in Cartesian coordinates) has components
$$
E = g_{11} = 1 + 4x^{2},\quad
F = g_{12} = 4xy,\quad
G = g_{22} = 1 + 4y^{2},
\tag{1}
$$
so if $(u, v)$ is a tangent vector at $(x, y)$, its length (squared) with respect to the paraboloid Riemannian metric is
\begin{align*}
Eu^{2} + 2Fuv + Gv^{2}
  &= (u^{2} + v^{2}) + 4(xu + yv)^{2} && \text{by (1)} \\
  &\leq (u^{2} + v^{2}) + 4(u^{2} + v^{2})(x^{2} + y^{2}) && \text{Cauchy-Schwarz} \\
  &= (u^{2} + v^{2})\bigl(1 + 4(x^{2} + y^{2})\bigr) && \\
  & \leq 5(u^{2} + v^{2}) && x^{2} + y^{2} \leq 1,
\end{align*}
i.e., five times the Euclidean magnitude (squared). If $p$ and $q$ are arbitrary points of the disk, the $d$-length of the line segment joining $p$ and $q$ is at most $\sqrt{5}$ times the Euclidean length $d'(p, q)$ of the segment, and this is no smaller than the geodesic $d$-distance from $p$ to $q$.
Whichever argument you prefer,
$$
d'(p, q) \leq d(p, q) \leq \sqrt{5}\, d'(p, q)\quad\text{for all $p$, $q$ in $M$.}
$$
This shows the two metrics are quasi-isometric.
(As a technical note, the closed unit disk isn't a Riemannian manifold, but a Riemannian manifold-with-boundary. Arguably, a "better" example of the theorem "any diffeomorphism of compact Riemannian manifolds is a quasi-isometry" would be an axial scaling that sends a sphere to an ellipsoid. Still, the closed disk computations should help convey the flavor of the definition.)
