The phrase "degrees of freedom" refers to the number of independent real number valued parameters (as does the more formal term "dimension").
So, for example, each orientation preserving isometry of $n$-dimensional Euclidean space $E^n$ is a composition $T \circ O$ of a unique orthogonal transformation $O$ followed by a unique translation $T$, and these two factors are independent of each other. Translations are parameterized by vectors which have $n$ degrees of freedom, and rotations of $E^n$ are parameterized by orthogonal $n \times n$ matrices which have $\frac{(n-1)n}{2}$ degrees of freedom. Since the degrees of freedom of translations and of rotations are independent of each other, together they have $n + \frac{(n-1)n}{2}=\frac{n(n+1)}{2}$ degrees of freedom.
Now let's bring in reflections. Let me use $R_n$ to denote the reflection in the coordinate plane $x_n=0$. A general orientation reversing isometry can be expressed uniquely in the form $T \circ O \circ R_n$ where $T,O$ are independently chosen translation and orthogonal transformation. It follows that a general isometry can be written uniquely in the form $T \circ O \circ (R_n)^e$ where we independently choose three things: the translation $T$ with $n$ degrees of freedom; the orthogonal rotation $O$ with $\frac{(n-1)n}{2}$ degrees of freedom; and the exponent $e \in \{0,1\}$.
The choice of the exponent $e$ is discrete --- either $0$ or $1$ --- and this does not represent a "degree of freedom" because it is not a real number valued parameter.
By the way, what makes the calculations in my answer work correctly is the fact that the decomposition $T \circ O \circ (R_n)^e$ is uniquely determined by the isometry. In your question, where you use the fact that every isometry can be written as a composition of reflections, that composition is not unique, and so using it to count degrees of freedom can lead to errors.
Just as an example, picking any even number $n \ge 2$, every translation of the line can be written as a product of $n$ reflections. Since a reflection of the line has $1$ degree of freedom (namely, the reflection point), this would seem to be a proof that the translations of the line have dimension $n$. Since $n$ is arbitrary, there is an error in this argument... which I will leave you to ponder.
I am not a mathematician, and I do not normally do proofs, as I only use math as a tool to find solutions to existing problems. So, rather than show the proof, I show how to construct the actual rotation matrix and translation vector for arbitrary 3D point triplets with positive pairwise distances.
(I hope one of the mathematicians here can show how to condense this into a proper proof, as a separate answer.)
When you have three points
$$\bbox{
\vec{p}_1 = \left [ \begin{matrix} x_1 \\ y_1 \\ z_1 \end{matrix} \right ]
}, \quad \bbox{
\vec{p}_2 = \left [ \begin{matrix} x_2 \\ y_2 \\ z_2 \end{matrix} \right ]
}, \quad \bbox{
\vec{p}_3 = \left [ \begin{matrix} x_3 \\ y_3 \\ z_3 \end{matrix} \right ]
}$$
with positive pairwise distances
$$\begin{aligned}
d_{12} &= \left\lVert\vec{p}_2 - \vec{p}_1\right\rVert = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \\
d_{13} &= \left\lVert\vec{p}_3 - \vec{p}_1\right\rVert = \sqrt{(x_3 - x_1)^2 + (y_3 - y_1)^2 + (z_3 - z_1)^2} \\
d_{23} &= \left\lVert\vec{p}_3 - \vec{p}_2\right\rVert = \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2 + (z_3 - z_2)^2} \\
\end{aligned}$$
you can construct a coordinate system where $\vec{p}_1$ is at origin, $\vec{p}_2$ is on the positive $x$ axis, and $\vec{p}_3$ is on the $xy$ plane on the positive $y$ side; i.e.
$$\bbox{
\vec{p}_1^\prime = \left [ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right ]
}, \quad \bbox{
\vec{p}_2^\prime = \left [ \begin{matrix} d_{12} \\ 0 \\ 0 \end{matrix} \right ]
}, \quad \bbox{
\vec{p}_3^\prime = \left [ \begin{matrix} u \\ v \\ 0 \end{matrix} \right ]
}$$
where
$$\bbox{\left\lbrace \; \begin{aligned}
u^2 + v^2 &= d_{13}^2 \\
(d_{12} - u)^2 + v^2 &= d_{23}^2
\end{aligned}\right .} \quad \iff \quad \bbox{ \left\lbrace \; \begin{aligned}
\displaystyle u &= \frac{d_{12}^2 + d_{13}^2 - d_{23}^2}{2 d_{12}} \\
\displaystyle v &= \sqrt{d_{13}^2 - u^2} \\
\end{aligned} \right .}$$
We can describe this transformation using an orthonormal matrix $\mathbf{M}$ and translation vector $\vec{t}$ as
$$\bbox{ \vec{p}^\prime = \mathbf{M} \left(\vec{p} - \vec{p}_1 \right) = \mathbf{M}\vec{p} + \vec{t} }$$
where
$$\bbox{ \vec{t} = -\mathbf{M} \vec{p}_1 }$$
and
$$\bbox{ \mathbf{M} = \left [ \begin{matrix}
m_{11} & m_{12} & m_{13} \\
m_{21} & m_{22} & m_{23} \\
m_{31} & m_{32} & m_{33} \end{matrix} \right ]} , \quad
\bbox{ \hat{u} = \left [ \begin{matrix} m_{11} \\ m_{12} \\ m_{13} \end{matrix} \right ]} , \quad \bbox{ \hat{v} = \left [ \begin{matrix} m_{21} \\ m_{22} \\ m_{23} \end{matrix} \right ]} , \quad \bbox{
\left [ \begin{matrix} m_{31} \\ m_{32} \\ m_{33} \end{matrix} \right ] = \hat{u} \times \hat{v} }$$
with
$$\bbox{ \left\lVert\hat{u}\right\rVert = \left\lVert\hat{v}\right\rVert = 1 },
\quad
\bbox{ \hat{u} \cdot \hat{v} = 0 }, \quad
\bbox{ \hat{u} = \frac{\vec{p}_2 - \vec{p}_1}{\left\lVert\vec{p}_2 - \vec{p}_1\right\rVert } }, \quad
\bbox{ \hat{v} = \frac{\vec{p}_3 - \vec{p}_1 - \hat{u} \left( \hat{u} \cdot \left ( \vec{p}_3 - \vec{p}_1 \right ) \right )}{\left\lVert \vec{p}_3 - \vec{p}_1 - \hat{u} \left( \hat{u} \cdot \left ( \vec{p}_3 - \vec{p}_1 \right ) \right ) \right\rVert} }$$
If we examine the matrix $\mathbf{M}$, we observe that it is orthonormal with determinant 1, and therefore a pure rotation matrix.
This means that if we have two such point triplets $\vec{p}_i$ and $\vec{q}_i$ with the same pairwise distance, and their corresponding rotation matrices $\mathbf{M}_1$ and $\mathbf{M}_2$, we find that
$$\mathbf{M}_1 ( \vec{p}_i - \vec{p}_1 ) = \mathbf{M}_2 ( \vec{q}_i - \vec{q}_1 )$$
because their transformed coordinates ($(0, 0, 0)$, $(d_{12}, 0, 0)$, and $(u, v, 0)$) are the same if their pairwise distances are the same. Reordering the terms we see that
$$\vec{p}_i = \mathbf{M}_1^{-1} \mathbf{M}_2 \vec{q}_i + \vec{p}_1 - \mathbf{M}_1^{-1} \mathbf{M}_2 \vec{q}_1$$
which can be written as
$$\vec{p}_i = \mathbf{R} \vec{q}_i + \vec{r}$$
where
$$\bbox{ \mathbf{R} = \mathbf{M}_1^{-1} \mathbf{M}_2 }, \quad
\bbox{ \vec{r} = \vec{p}_1 - \mathbf{R} \vec{q}_1 }$$
Thus, we have constructed the rotation matrix $\mathbf{R}$ and translation vector $\vec{r}$. Hopefully, its existence is proof enough.
Note that the above case is for the rotation only. However, if you change the third row vector of $\mathbf{M}$ into $\hat{v} \times \hat{u} = -\hat{u} \times \hat{v}$, its determinant will be -1, so it will be an "improper rotation matrix": a reflection.
Best Answer
They're not the only ones: Consider $$ (x,y) \mapsto (x,-y) \mapsto (x+1,-y). $$ This is not a reflection about a line, nor a translation, nor a rotation. However, it is a composition of a reflection and a translation. The answer should be that translations, reflections, and rotations GENERATE the group of all isometries of the plane, i.e. every isometry is a composition of finitely many those. (Auxiliary question: How many? Do we ever need more than two?)
("Glide reflection" is a term I've seen used for isometries like the one displayed above. I don't know what degree of standardness that term has.)
In three dimensions, you could also rotate through a $1^\circ$ angle and then translate along the axis of rotation. By contrast, in two dimensions, if you rotate and then translate, what you get is just a rotation about a different center.