Let $S \subset {\Bbb R}^3$ be a (sufficiently) smooth surface, and let $\sigma (t)$, $\tau(t)$ be two smooth curves in $S$. Suppose $\sigma(t)$ and $\tau(t)$ both pass through the point $p \in S$; without loss of generality we can take $\sigma(0) = \tau(0) = p$. Since $\sigma(t)$ and $\tau(t)$ are also curves in $\Bbb R^3$, their tangent vector fields $\sigma'(t)$, $\tau'(t)$ lie in $T \Bbb R^3$, the tangent bundle of $\Bbb R^3$. As such, we can take the inner product of $\sigma'(t)$ and $\tau'(t)$ at any point such as $p$ through which they both pass by exploiting the Euclidean inner product structure $\langle \cdot, \cdot \rangle_{\Bbb R^3}$, viz. by taking for example $\langle \sigma'(0), \tau'(0) \rangle_{\Bbb R^3}$; we can also obtain the magnitudes of these tangent vectors for any value of $t$ in a similar fashion, by taking e.g. $\Vert \sigma'(t) \Vert_{\Bbb R^3} = \sqrt{\langle \sigma'(t), \sigma'(t) \rangle_{\Bbb R^3}}$ with the analogous expression holding for $\tau(t)$. And, having the norms of these tangent vectors, we can in principle compute the lenths if curve segments such as $\sigma(t)$, $t_1 \le t_2$, via the formula
$l(\sigma, t_1, t_2) = \int_{t_1}^{t_2} \Vert \sigma'(t) \Vert_{\Bbb R^3} dt; \tag{1}$
and again, the corresponding formula holds for $\tau(t)$. All these quantities are defined with reference to $\Bbb R^3$, since the all invoke $\langle \cdot, \cdot \rangle_{\Bbb R^3}$ in their definitions, and indeed yield geometrical information about $\sigma(t)$, $\tau(t)$ which in no way requires knowledge of the surface $S$; we merely exploit the fact that $\sigma(t)$, $\tau(t)$ are curves in the ambient space $\Bbb R^3$.
On the other hand, we may also define a tensor field $I: TS \times TS \to \Bbb R$ by taking
$I(\sigma'(0), \tau'(0)) = \langle \sigma'(0), \tau'(0) \rangle_{\Bbb R^3} \tag{2}$
for tangent vectors $\sigma'(0), \tau'(0) \in T_pS$, allowing $p$ to vary over $S$ and adjusting $\sigma(t)$, $\tau(t)$ accordingly so that we always have $\sigma(0) = \tau(0) = p$ while the curves remain in $S$. Such a construction allows the definition of $I$ to be extended to all of $TS$. Once $I$ has been so defined, admittedly in terms of $\langle \cdot, \cdot \rangle_{\Bbb R^3}$, it may be viewed as a tensor field on $S$ without further reference to $\Bbb R^3$; all metric properties of $S$ may now be defined solely in terms of $I$: we have
$\Vert \sigma'(0) \Vert_S = \sqrt{I(\sigma'(0), \sigma'(0))}, \tag{3}$
$l(\sigma, t_1, t_2) = \int_{t_1}^{t_2} \Vert \sigma'(t) \Vert_S dt, \tag{4}$
and we can define an inner product on $TS$ via
$\langle \sigma'(0), \tau'(0) \rangle_S = I(\sigma'(0), \tau'(0)). \tag{5}$
We may now consider $I$ as a structure defined on $TS$ alone. Doing so, we obtain all metric properties of $S$ without need to again refer to $\Bbb R^3$.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
$\newcommand{\Reals}{\mathbf{R}}\newcommand{\dd}{\partial}$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 $\Reals^{3}$.
Best Answer
What you're looking for is known as a Riemannian metric. One way of phrasing it is as follows. (Lee talks about these later in a slightly different way of phrasing later in his book.)
A Riemannian metric on a smooth manifold $M$ is a choice of inner product $g_p$ on each tangent space $T_pM$ that varies smoothly, in the sense that given any two smooth vector fields $X$ and $Y$ on $M$, the function $g(X,Y) = p \mapsto g_p(X_p,Y_p)$ is smooth.
Now let's start with an embedding $f: M \hookrightarrow \Bbb R^n$. The first fundamental form gives us a Riemannian metric on $M$ by setting $g_p(x,y) = Df(x) \cdot Df(y)$. To see that this is smoothly varying as defined above, consider the maps $TM \to \Bbb R^n \times \Bbb R^n \to \Bbb R^n, (p,v) \mapsto (f(p), Df(v)) \mapsto Df(v)$. Write a smooth vector field $X$ as $p \mapsto (p,X(p))$. Then the map $M \to \Bbb R^n, p \mapsto Df(X(p))$ is smooth, and so is the inner product map $\Bbb R^n \times \Bbb R^n \to \Bbb R, (x,y) \mapsto x\cdot y$. It is readily seen that $g(X,Y) = Df(X(p)) \cdot Df(Y(p))$ and that this latter map is smooth, as desired.
I should also say that a smooth manifold does not automatically come with a Riemannian metric, which is much more structure.