Here is one way to get the conjugacy classes of $D_n$ and irreducible representations over $\mathbf{C}$.
Setup
First I will fix some notation. A presentation of $D_n$ is $\langle r, s\mid r^n = s^2 = 1, sr=r^{-1}s\rangle$, which means we can pin down $D_n$ as a group of rotations $\{ 1, r, r^2\ldots r^{n-1}\}$ together with a bunch of reflections $\{ s, sr, sr^2\ldots sr^{n-1}\}$. (In what follows I will frequently be sloppy and implicitly use the identifications $r^{-i} = (r^i)^{-1} = r^{n-i}$.)
Analysis of conjugacy classes of $D_n$
To figure out the conjugacy classes, we can just use brute force. Every element is $r^i$ or $sr^i$ for $0\leq i < n$, so it's not too hard just to write down every possible conjugation:
\begin{array}{rlclcl}
\text{Conjugate} &r^i &\text{by}&r^j &:&(r^j) r^i(r^{-j})=r^i\\
&r^i &\text{by}&sr^j&:&(sr^j)r^i(r^{-j}s)= sr^is=r^{-i}\\
&sr^i&\text{by}&r^j &:&(r^j) sr^i(r^{-j}) = sr^{-j}r^ir^{-j} = sr^{i-2j}\\
&sr^i&\text{by}&sr^j&:&(sr^j)sr^i(r^{-j}s) = r^{i-2j}s=sr^{2j-i}\\
\end{array}
Conjugacy classes of rotations
The first two rows tell us that the set of rotations decomposes into inverse pairs $r^i$ and $(r^{i})^{-1}$, i.e. the classes $\{1\}, \{r, r^{n-1}\}, \{r^2, r^{n-2}\},\ldots$
Counting these, there are $\frac{n}{2}+1$ when $n$ is even (note that $r^{n/2}$ is its own inverse) and $\frac{n+1}{2}$ when $n$ is odd.
Conjugacy classes of reflections
Now observe from the third and fourth lines of the table that $sr$ is conjugate to $sr^3, sr^5, \ldots$ while $s$ is conjugate to $sr^2, sr^4,\ldots$ and these two sets are disjoint if $n$ is even. However, $sr$ is conjugate to $sr^{n-1}$ (via $r$) so if $n$ is odd, all the nontrivial reflections are in one conjugacy class. (You said you already knew this but I'm putting it here for completeness.)
Together, this brings us to the total number of conjugacy classes of $D_n$:
\begin{array}{rl}
\left(\frac{n}{2}+1\right)+2 = \color{#090}{\frac{n}{2}+3}&\text{for }n\text{ even}\\
\left(\frac{n+1}{2}\right)+1 = \color{#090}{\frac{n+3}{2}}&\text{for }n\text{ odd.}
\end{array}
Analysis of irreducible representations of $D_n$
One dimensional irreducibles
The commutators of $D_n$ look like $[r^i, sr^j]$ or the inverse of such, and
$$[r^i, sr^j] = r^{-i}(sr^j)r^i(sr^j) = sr^{2i+j}sr^j = (r^i)^2$$ so the commutators generate the subgroup of squares of rotations. This means $G/[G,G]$ has order 2 if $n$ is odd (since all rotations are squares) or order 4 if $n$ is even (since only half the rotations are squares). Now you can use your fact #4, which tells us that we have precisely 2 ($n$ odd) or 4 ($n$ even) irreps of dim. 1 obtained from pulling back those from $G/[G,G]$.
Other irreducibles
This is related to your item #5. We can define some 2-dim'l representations over $\mathbf{R}$, namely
\begin{array}{ccc}
r&\mapsto&\pmatrix{\cos(2\pi k/n)&-\sin(2\pi k/n)\\
\sin(2\pi k/n)&\cos(2\pi k/n)} \\
s&\mapsto&\pmatrix{0&1\\1&0}
\end{array}
for $0\leq k \leq \lfloor \frac{n}{2}\rfloor$. We would like to know if these rep'ns are irreducible if we consider them as matrices over $\mathbf{C}$.
They are reducible if $k=0$ or $k = n/2$ (can you decompose them?). If $k$ is different from $0$ or $n/2$, a quick computation shows that the matrix for $r$ has distinct complex eigenvalues $\pm e^{2\pi ki/n}$ with corresponding eigenvectors $\pmatrix{1\\-i}$ and $\pmatrix{1\\i}$. The spans of each e-vector are the only candidates for invariant subspaces, but the matrix for $s$ interchanges the two eigenspaces, so there are no invariant subspaces and thus these repn's are irreducible.
Final count
We have 2-dim'l irreps for each integer $1\leq k < \frac{n}{2}$, specifically we have $\frac{n}{2}-1$ for $n$ even and $\frac{n-1}{2}$ for $n$ odd. If we count these with the 1-dim'l irreps, we have
\begin{array}{rl}
\left(\frac{n}{2}-1\right)+4 = \color{#090}{\frac{n}{2}+3}&\text{for }n\text{ even}\\
\left(\frac{n-1}{2}\right)+2 = \color{#090}{\frac{n+3}{2}}&\text{for }n\text{ odd.}
\end{array}
which matches up with the number of conjugacy classes, so we must be done by your fact #1. (Also, we can verify that the sum of squares of the dimensions of the irreps is $2n$ in both cases.)
Some stuff I scrounged up. First, an easier result: an element $g$ of a finite group $G$ is real if it is conjugate to $g^{-1}$, or equivalently if the character of $g$ is always real in every representation. So the following conditions are equivalent:
- Every element of $G$ is conjugate to its inverse.
- Every character of $G$ is real.
- Every finite-dimensional representation of $G$ is either real or quaternionic.
Such groups are called ambivalent. This is a necessary condition for all representations being real; the symmetric groups obviously satisfy it, but I'm not sure if all finite Coxeter groups do (edit: they do, see the comments). However, it turns out to be satisfied by all Weyl groups.
The stronger condition that every representation is real is called being totally orthogonal; apparently it's known (see the link) that any such group must be generated by involutions. It doesn't seem like necessary and sufficient conditions are known in general.
Finally, the Weyl groups satisfy an even stronger condition which implies total orthogonality, that every representation is realizable over $\mathbb{Q}$: this is claimed in Humphreys in Section 8.10 and the citation is to Benard's On the Schur indices of characters of the exceptional Weyl groups. Such a group must in particular satisfy the property that every character is integer-valued and this condition is called being rational.
Best Answer
There are two concepts here. The first is in what sense are the irreducible representations the building blocks of all representations. This, as mentioned in the comments, is because every representation over $\mathbb C$ is isomorphic to a direct sum of irreducible representations.
The second, which you are alluding to, is whether, given the $2$-dimensional representation, you need the others. What you seem to be driving at is that the $2$-dimensional representation is faithful, meaning the homomorphism to the matrix group is injective. Indeed, the $2$-dimensional representation is faithful, and any faithful representation of $D_4$ must have that $2$-dimensional representation as a summand. In that sense you do not need the other representations to define your group, and the other representations cannot be used, in that you always need the $2$-dimensional representation.
A finite group has a faithful, irreducible representation if and only if its centre is cyclic. So for the group $C_2\times C_2$ you cannot find a faithful irreducible representation, and you would need the sum of two of them to produce an isomorphic copy of your group. In general, if the centre needs $n$ elements to generate it, one needs $n$ irreducible representations to be faithful (and of course, you must choose your representations wisely to obtain a faithful representation).