You can parametrize your curves as $c_1(t)=(t,t)$ and $c_2(t)=(t,-t)$ (first coordinate: $u$). You are searching for the angle between the tangent vectors $v=(1,1)$ and $w=(1,-1)$ (tangent vectors along $c_1$ and $c_2$: note that they do not depend on $t$) using the metric $g=du^2+dv^2$ with components $g_{ij}(u,v)=\delta_{ij}$ for $i,j=1,2$.
Then the angle $\theta$ is s.t. $\cos\theta=\frac{\langle v,w\rangle}{\|v\| \|w\|}$,
with $\langle v,w\rangle=g_{ij}v_iw_j$ and $\|v\|=\sqrt{g_{ij}v_iv_j}$ (and similarly for $\|w\|$). We sum over repeated indices.
Let $\sigma: U\subseteq \mathbb R^2\rightarrow \mathbb R^3$ $(u,v)\mapsto \sigma(u,v)=(x(u,v),y(u,v),z(u,v))$ parametrize a 2 dimensional surface in $\mathbb R^3$ by coordinates $(u,v)$.
With
$$I(X,Y):=\langle X,Y\rangle:=\sum_{i,j=1}^2 a_ib_j\langle \sigma_i,\sigma_j\rangle $$
we denote its first fundamental form, computed for all tangent vectors
$$X=a_1\sigma_1+a_2\sigma_2, $$
$$Y=b_1\sigma_1+b_2\sigma_2, $$
at any given point of the surface.
We introduced the notation $\sigma_1:=\sigma_u=\frac{\partial \sigma}{\partial u}$ and $\sigma_2:=\sigma_v=\frac{\partial \sigma}{\partial v}$ for sake of clarity.
A translation
$$\sigma'(u,v):=\sigma(u,v)+b, $$
with $b\in\mathbb R$ has no effect on the tangent vectors $\sigma_u$ and $\sigma_v$, i.e.
$$\sigma'_u=\sigma_u, $$
$$\sigma'_v=\sigma_v, $$
and the first invariant form is unaffected by translations.
This follows immediatly by computing $\frac{\partial \sigma'}{\partial u}$ and $\frac{\partial \sigma'}{\partial v}$.
- Orthogonal transformations
By definition of orthogonal transformation (represented by a matrix $O$, once we choose a basis in $\mathbb R^3$)
$$I(OX,OY):=\langle OX,OY\rangle=\langle O^TOX,Y\rangle=\langle X,Y\rangle=I(X,Y),$$
as $O^TO=I$. The first invariant form is then invariant under orthogonal transformations. In the reference presented in the OP this fact is summarized under "...and $P$ preserves the dot product".
A dilation
$$\sigma'(u,v):=\alpha\sigma(u,v), $$
with $\alpha\in\mathbb R^{+}$ is such that
$$\sigma'_u=\alpha\sigma_u, $$
$$\sigma'_v=\alpha\sigma_v; $$
we arrive at
$$I(X',Y')=\sum_{i,j=1}^2 a_ib_j\langle \sigma'_i,\sigma'_j\rangle =\sum_{i,j=1}^2 a_ib_j\langle \alpha\sigma_i,\alpha\sigma_j\rangle=a^2I(X,Y),$$
i.e. applying a dilation by $\alpha$ the first fundamental form is multiplied by $\alpha^2$.
Best Answer
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\dd}{\partial}$If you have a surface embedded in $\Reals^{3}$, you can (as you note) use the "ambient" Euclidean inner product to take dot products of tangent vectors. However, that's mild overkill; in order to take dot products of tangent vectors to a surface, all you really need is the first fundamental form.
If $$ E = \left\langle\frac{\dd q}{\dd u}, \frac{\dd q}{\dd u}\right\rangle,\quad F = \left\langle\frac{\dd q}{\dd u}, \frac{\dd q}{\dd v}\right\rangle,\quad G = \left\langle\frac{\dd q}{\dd v}, \frac{\dd q}{\dd v}\right\rangle, $$ and if $$ v = a_{1}\, \frac{\dd q}{\dd u} + a_{2}\, \frac{\dd q}{\dd v},\qquad w = b_{1}\, \frac{\dd q}{\dd u} + b_{2}\, \frac{\dd q}{\dd v}, $$ i.e., if $v = (a_{1}, a_{2})$ and $w = (b_{1}, b_{2})$ with respect to the coordinate basis fields, then $$ \langle v, w\rangle = a_{1} b_{1} E + (a_{1} b_{2} + a_{2} b_{1}) F + a_{2} b_{2} G. $$ That's the content of the first paragraph of your excerpt. The second paragraph asserts that $\langle v, w\rangle$ is independent of the choice of local coordinates. This is in some sense "obvious" if you know you have a surface embedded in $\Reals^{3}$, but it's surprising if you think of the first fundamental form as "extra structure" imposed on an open set in $\Reals^{2}$, i.e., in a coordinate neighborhood.
Remarkably, there's non-trivial "intrinsic" geometry of a surface captured by its first fundamental form, even though the first fundamental form doesn't uniquely determine an embedding of the surface in $\Reals^{3}$.