In order to determine the finite SCT from its infinitesimal version, we need to solve for the integral curves of the special conformal killing field $X$ defined by
\begin{align}
X(x) = 2(b\cdot x) x - x^2 b.
\end{align}
I explain why this is equivalent to "integrating" the infinitesimal transformation below. This means we need to solve the differential equation $X(x(t)) = \dot x(t)$ for the function $x$. Explicitly, this differential equation is
\begin{align}
\dot x = 2(b\cdot x) x - x^2 b.
\end{align}
This can be done with a trick, namely a certain change of variables. Define
\begin{align}
y = \frac{x}{x^2},
\end{align}
then the resulting differential equation satisfied by $y$ becomes simple;
\begin{align}
\dot y = -b.
\end{align}
I urge you to perform the algebra yourself to confirm this. It's kind of magic that it works if you ask me, and the change of variables is precisely an inversion, so I think there's something deeper going on here, but I'm not sure what it is. The solution to this equation is simply $y = y_0 -tb$, so we find that the original function $x$ satisifes
\begin{align}
\frac{x}{x^2} = \frac{x_0}{x_0^2} - tb.
\end{align}
In other words, we've turned a monstrous nonlinear system of ODEs into a simple algebraic equation. In fact, one can show that the algebraic eqution $x/x^2 = A$ has the solution $x = A/A^2$, from which it follows that the solution to our original problem is
\begin{align}
x(t) = \frac{x_0 - x_0^2(tb)}{1-2x_0\cdot(tb) + x_0^2(tb)^2},
\end{align}
as desired, since this is precisely the form of the "finite" SCT. Note that these only are local integral curves; the solution hits a singularity when $t$ is such that the denominator vanishes.
Why solve for integral curves?
If you're wondering what your original question has anything to do with solving the integral curves of the special conformal vector field I wrote down, then read on.
It helps to start with the concept of a flow.
Transforming points via flows.
Let a point $x\in\mathbb R^d$ be given, then for each $b\in\mathbb R^d$, we assume, at least in a neighborhood of that point, that there is a $\epsilon$-parameter family of transformations $\Phi_b(\epsilon):\mathbb R^d \to \mathbb R^d$ with $\epsilon\in [0,\bar\epsilon)$ such that $\Phi_b(\epsilon)(x)$ tells you what an SCT corresponding to the vector $b$ does to the point $x$ is you ``flow" in $\epsilon$. At $\epsilon = 0$, this flow just maps the point to itself;
\begin{align}
\Phi_b(0)(x) = x,
\end{align}
namely it starts at the identity. For $\epsilon >0$, the flow translates the point along a curve in $\mathbb R^d$. If you change $b$, then this corresponds to moving way from $x$ in a different initial direction under the flow.
Infinitesimal generator of a flow.
When we talk of an infinitesimal generator of such a transformation, we are talking about the term that generates the linear approximation for the flow in the parameter $\epsilon$. In other words, we expand
\begin{align}
\Phi_b(\epsilon)(x) = x + \epsilon G_b(x) + O(\epsilon^2),
\end{align}
and the vector field $G_b$ is called the infinitesimal generator of the flow. What you have pointed out in your question is that we know this infinitesimal generator;
\begin{align}
G_b(x) = 2(b\cdot x)x - x^2 b,
\end{align}
and we now want to reconstruct the whole flow simply by knowing this information corresponding to its linear approximation at every point. This is equivalent to solving some first order ordinary differential equations, which is why people often say we want to "integrate" the infinitesimal transformation to determine the finite one; integration is a perhaps somewhat archaic way of solving the corresponding differential equation.
Finding the whole flow given its generator.
Ok, so what differential equation do we solve? Well, note that the vector field $G_b$ is tangent to the curves generated by the flow by its very construction (we took a derivative with respect to $\epsilon$ with is the "velocity" of the curve generated by the flow), so the differential equation we want to solve is
\begin{align}
\dot x(\epsilon) = G_b(x(\epsilon)),
\end{align}
and we want to solve for $x(\epsilon)$. The solutions to this differential equation are referred to as integral curves of the vector field $G_b$.
Acknowledgements.
I didn't figure out the first part of this answer completely on my own. The idea for making the substitution $y=x/x^2$, which is really the crux of everything, came from here http://www.physicsforums.com/showthread.php?t=518316, namely from user Bill_K.
The idea for how to solve the algebraic equation $x/x^2 = A$ came from math.SE user @HansLundmark after I posted essentially your question in mathy language on math.SE here.
I should another math.SE user @Kirill solved for the integral curves in a totally different way in his answer to the question I posted.
Addendum.
How does one get $\dot y = -b$ from the change of variable $y=x/x^2$ as claimed in the first section? Well, let's calculate:
\begin{align}
\dot y
&= \frac{x^2\dot x - x(2x\cdot \dot x)}{(x^2)^2} \\
&= \frac{x^2(2(b\cdot x) x - x^2 b) - x(2x\cdot (2(b\cdot x) x - x^2 b))}{(x^2)^2} \\
&= \frac{2x^2(b\cdot x) x - (x^2)^2b - 4x^2(b\cdot x) x + 2x^2(b\cdot x)x}{(x^2)^2} \\
&= -\frac{(x^2 )^2b}{(x^2)^2} \\
&= -b
\end{align}
Magic!
Best Answer
The first equality for translations is true for any field in general. Since translations form an Abelian group, their irreducible representations are one-dimensional. Thus, under a translation, it is generally true for an irreducible field (for any spin) $\Phi'(x+a) = \Phi(x)$. Of course, the situation changes when the Lorentz group is considered. In this case, the field also transforms nontrivially and $F \neq Id$.
The second case with dilations is not generally true. It is only true for a class of fields called quasi-primary fields that transform as $\Phi'(\lambda x) = \lambda^{-\Delta} \Phi(x)$. However, in a CFT, it is a fact that the complete set of fields can be grouped into "conformal families", each with its representative primary field and its descendants. Transformation properties of the descendants can be derived from its corresponding primary field. Thus, it is often the case in CFTs, that one only discusses the transformation of primary fields. This is the reason Francesco chooses to discuss the example that he did rather than something else.