This is a famous example of a finite group of linear fractional transformations. These are transformations of
the form
$$
x\mapsto \frac{ax+b}{cx+d},
$$
wher $ad-bc\neq0$. The use of the term "linear transformation" here comes from the following. Imagine the set of lines in the plane through the origin.
Such a line $L$ is a 1-dimensional subspace, and thus generated by a single vector, say the vector $\vec{v}_L=(x_L,y_L)$. Because any non-zero scalar multiple of $\vec{v}_L$ generates the same subspace we can scale the $y$-coordinate to be equal to $1$, i.e. use $(x_L/y_L,1)$ as the generator
instead. This fails only, when $y_L=0$ (i.e. when the line is horizontal) in which case we scale $x_L$ to $1$ instead.
So the ratio $\lambda_L=x_L/y_L$ determines the line uniquely, and we can cover all the lines, if we define $\lambda=1/0=\infty$ for the horizontal line. The parameter $\lambda$ is thus the reciprocal of the slope of the line $L$.
Anyway, an invertible linear transformation $T$ from the plane to itself will map such a 1-dimensional subspace to another. Assume that the linear transformation amounts to multiplication (from the left side) by the matrix
$$
A=\left(\begin{array}{cc}a&b\\c&d\end{array}\right).
$$
Let's check what this does to the parameter $\lambda$. We see that $$T(\vec{v}_L)=A\left(\begin{array}{c}\lambda\\1\end{array}\right)=\left(\begin{array}{c}a\lambda+b\\c\lambda+d\end{array}\right).$$
The $\lambda$-parameter of the image $T(L)$ is thus $(a\lambda+b)/(c\lambda+d)$. It is also easy to see that the line with $\lambda=-d/c$ is mapped to the horizontal line with $\lambda=\infty$, and that horizontal line $\lambda=\infty$ is mapped to the line with $\lambda=a/c$ because $T(1,0)=(a,c)$. Getting warmer, right?
Now it is easy to check that composition
of planar linear transformations, known to correspond to a product of matrices, also corresponds to the composition of fractional linear transformations. In the language of groups this means that the function mapping the above matrix to the function $\lambda\mapsto (a\lambda+b)/(c\lambda+d)$ is a homomorphism of group.
But, there's a catch. Should $A=aI_2$ be a scalar matrix, that is $a=d, b=c=0$, then $(a\lambda+b)(c\lambda+d)=a\lambda/a=\lambda$ for all $\lambda$.
In other words, the scalar matrices map all those lines to themselves.
Geometrically this is obvious because the scalar linear transformations simply stretch things radially away from the origin. In the language of groups this means that the scalar matrices belong to the kernel of that homomorphism.
You may have already guessed where this is heading. The functions in your example homomorphic images of the matrices
$$
\begin{aligned}
A_1&=I_2\mapsto&[x\mapsto x],\\
A_2&=\left(\begin{array}{cc}0&1\\1&0\end{array}\right)\mapsto&[x\mapsto\frac1x],\\
A_3&=\left(\begin{array}{cc}-1&1\\0&1\end{array}\right)\mapsto&[x\mapsto 1-x],\\
A_4&=\left(\begin{array}{cc}0&1\\-1&1\end{array}\right)\mapsto&[x\mapsto\frac1{1-x}],\\
A_5&=\left(\begin{array}{cc}1&-1\\1&0\end{array}\right)\mapsto&[x\mapsto\frac{x-1}{x}],\\
A_6&=\left(\begin{array}{cc}1&0\\1&-1\end{array}\right)\mapsto&[x\mapsto\frac x{x-1}].\\
\end{aligned}
$$
The above catch shows here as follows. The above set of six matrices does not form a group under matrix multiplication. For example you easily see that $A_4^2=-A_5$. But, the matrices $A_5$ and $-A_5$ give rise to the same linear fractional transformation. Furthermore, the set of twelve matrices
$$G=\{\pm A_i\mid i=1,2,3,4,5,6\}$$
does form a group under matrix multiplications. Leaving it to you to verify that.
Summary: This specific set of functions forms a group because they come from a finite group of invertible $2\times2$-matrices, and there composition faithfully matches the product of those matrices (or linear transformations).
In spite of what that Wikipedia article may be implying these groups are very much of interest over fields other than a subfield of the complex numbers, but let's leave that for some other day :-)
Best Answer
The main ingredient in my answer is the following fact:
Its proof isn't directly related to this question so I won't put it right here. But you can find it in the answer to this question. Due to this fact we can talk about the $\mathbb R$ as about the disjoint union $\coprod\limits_{i=1}^\infty R_i$ where all of the $R_i$'s are the copies of $\mathbb{R}$.
Now we can define two functions $\mathbb{R} \to \mathbb{R}$: $F$ and $G$. $F$ sends $R_i$ due to the map $g_i$ (namely, $F|_{R_i} = g_i \circ T_i$, where $T_i$ is an arbitrary bijection from $R_i$ to $\mathbb R$) and $G$ works by the following rule: it's defined as a function $\coprod\limits_{i=1}^\infty R_i \to \coprod\limits_{i=1}^\infty R_i$ which sends $T_{i-1}^{-1}(a)$ to $T_i^{-1}(a)$ for all $a$ in $\mathbb R$. It's a well-defined map since all $T_i$'s are bijections.
The rest of the proof is quite elementary: you can easily check that $g_i = F \circ G^{i-1} \circ {T_1}^{-1}$. So your finite set of functions $f_i$ is just $\{F, G, T_1\}$.