First, a recap of the usual Jordan form. Consider a complex vector space $V$ and a linear map $\l$ on it, such that $V$ is a $\C[\l]$-module in the obvious way. Since $\C$ is algebraically closed, the elementary factors of $V$ have the form $(\l-\a)^k$, so
$$V \cong \bigoplus_i \C[\l]/(\l-\a_i)^{k_i}.$$
The point of Jordan form is that each $\l$-span $\C[\l]/(\l-\a_i)^{k_i}$ admits a basis
$$\{1,\ \l-\a_i\,, \ldots,\ (\l-\a_i)^{k_i-1}\}$$
that satisfies the simple relations
$$\l(\l-\a_i)^k = \a_i(\l-\a_i)^k+(\l-\a_i)^{k+1}.$$
Thus the action of $\l$ on this $\l$-span can be represented by the matrix
$$\begin{pmatrix} 1 & \a_i & & & \\ & 1 & \a_i & & \\ & & \ddots & & \\ & & & 1 & \a_i \\ & & & & 1 \end{pmatrix}$$
As a whole, the matrix of $\l$ can be represented as a direct sum of these Jordan blocks.
It turns out that we can get something close to Jordan form when working purely in $\R$. So now suppose $V$ is a real vector space and the linear map $\l$ induces an $\R[\l]$-module structure. The key fact is that all irreducibles in $\R[\l]$ are either linears or quadratics. Therefore, in the elementary factor decomposition of $V$, each direct summand is either
$$\R[\l]/(\l-\a)^k \quad\text{or}\quad \R[\l]/(\l^2-\a\l-\b)^k, \quad\text{$\l^2-\a\l-\b$ irreducible.}$$
If the former, we already know from the above discussion that there is a nice matrix form on that $\l$-span. So it is the latter case that is new.
This is where the magic of complexification comes in. Suppose $V$ is the $\l$-span $\R[\l]/(\l^2-\a\l-\b)^k$. The idea is that we make $V$ into a complex vector space
$$V^\C \cong \C[\l]/(\l^2-\a\l-\b)^k,$$
and by CRT $V^\C$ splits into two 'nice' $\l$-spans $\C[\l]/(\l-\g)^k$ and $\C[\l]/(\l-\ol\g)^k$, where $\g,\ol\gamma$ are the roots to $\l^2-\a\l-\b$. By the above discussion we have two nice sets of complex bases $\\{\v_1,\ldots,\v_k\\}$ and $\\{\w_1,\ldots,\w_k\\}$ for each of the summands, which we then use to obtain a real basis $\\{v_1',\ldots,v_{2k}'\\}$ under which $\l$ has a nice matrix form.
Usually $V^\C$ is straightaway defined as the scalar extension $\C\otimes_\R V$, but it might be more illuminating to list out the properties that we want $V^\C$ to satisfy, and then prove $\C\otimes_\R V$ is indeed a model (which I'll leave as an exercise). So, $V^\C$ should satisfy these very reasonable properties:
Elements of $V^\C$ can be written in the form $v+wi$ where $v,w\in V$. There is an embedding $V\hookrightarrow V^\C$ mapping $v\mapsto v+0i$.
Addition is defined in the obious way. Scalar multiplication is defined by the formula
$$(a+bi)(v+wi) = (av-bw)+(aw+bv)i, \quad a,b\in\R.$$
A linear map $\l$ on $V$ extends to a map on $V^\C$ by the formula
$$\l(v+wi) = \l v + (\l w)i.$$
Now we establish some basic facts about $V^\C$.
Proposition 1. A basis $\\{v_1,\ldots,v_n\\}$ is also a basis for $V^\C$.
Proof. If $\sum_k (a_k+b_ki)(v_k+0i) = 0$ then $(\sum_k a_kv_k) + (\sum_k b_kv_k)i = 0$, so $\sum_k a_kv_k=\sum_k b_kv_k=0$ and thus $a_k=b_k=0$ for all $k$. Thus linear independence is preserved.
If $v,w\in V$ then $v=\sum_k a_kv_k$ and $w=\sum_k b_kv_k$, so
$$\begin{align*} v+wi &= \sum_k a_kv_k + \left(\sum_k b_kv_k\right)i \\ &= \sum_k a_kv_k + \left(\sum_k b_ki\cdot v_k\right) \\ &= \sum_k (a_k+b_ki)v_k, \end{align*}$$
so $v_1,\ldots,v_n$ span $V^\C$ as well. $\qed$
Proposition 2. The extension of $\l:V\to V$ to $V^\C$ as defined above is automatically $\C$-linear. Furthermore it satisfies
$$\ol{\l(v+wi)} = \l(\ol{v+wi}) = \l(v-wi).$$
Proof. Additivity is trivial. And we have
$$\begin{align*} (a+bi)\l(v+wi) &= (a+bi)\bigl(\l(v)+\l(w)i\bigr) \\ &= \bigl(a\l(v)-b\l(w)\bigr)+\bigl(a\l(w)+b\l(v)\bigr)i \\ &= \l(av-bw)+\l(aw+bv)i \\ &= \l\bigl((av-bw)+(aw+bv)i\bigr) \\ &= \l\bigl((a+bi)(v+wi)\bigr). \end{align*}$$
As for the second claim, let $v_1,\ldots,v_n$ be a basis for $V$. Since elements of $V$ are self-conjugate we have
$$\ol{\l v_i} = \l v_i = \l\ol{v_i}.$$
By proposition 1, the $v_i$'s also form a basis of $V^\C$, so it suffices to prove that linear combinations preserve the equality between the first and third terms. This is pure calculation:
$$\begin{align*} \ol{\l\left(\sum_k (a_k+b_ki)v_k\right)} &= \ol{\sum_k (a_k+b_ki)\l v_i} \\ &= \sum_k \ol{a_k+b_ki}\cdot\ol{\l v_i} \\ &= \sum_k \ol{a_k+b_ki}\ \l\ol{v_i} \\ &= \l\left(\ol{\sum_k (a_k+b_ki)v_k}\right).\ \qed \end{align*}$$
Proposition 1 is very important, because if $M$ is the matrix representing $\l$ in $V$, then we can also use $M$ to represent $\l$ in $V^\C$! Thus, the characteristic polynomial of $\l$ in $V^\C$ is also $(\l^2-\a\l-\b)^k$, but this time it splits into linear factors. Therefore,
$$V^\C \cong \C[\l]/(\l^2-\a\l-\b)^k \cong \C[\l]/(\l-\g)^k \oplus \C[\l]/(\l-\ol\g)^k.$$
Put in words, the complexification of the real $\l$-span splits into two 'nice' complex $\l$-spans. Thus we have two separate (complex) bases $\\{\v_1,\ldots,\v_n\\}$ and $\\{\w_1,\ldots,\w_n\\}$ such that
$$\begin{align*} \l \v_k &= \g\,\v_k + \v_{k+1}, \\ \l \w_k &= \ol\g\,\w_k + \w_{k+1}. \end{align*}$$
But conjugating the first equation yields
$$\l\ol{\v_k} = \ol\gamma\,\ol{\v_k} + \ol{\v_{k+1}},$$
meaning that given the basis $\\{\v_k\\}$, we can in fact choose $\w_k=\ol{\v_k}$! Now define the real vectors
$$v_k' = {1\over2}(\v_k+\ol{\v_k}),\quad w_k' = {1\over2}(\v_k-\ol{\v_k}).$$
Note that $v_k'$ and $w_k'$ are simply the real and imaginary parts of $\v_k$. The corresponding relations matrix has the block form
$$\begin{pmatrix} 1/2 & 1/2 & & \\ 1/2 & -1/2 & & \\ & & 1/2 & 1/2 & \\ & & 1/2 & -1/2 & \\ &&&& \ddots \end{pmatrix},$$
which has nonzero determinant and is hence invertible. Thus the $v_k'$'s and $w_k'$'s together form a real basis for $V^\C$. Now let $\g=a+bi$. Since $\v_k = v_k'+w_k'i$, we can deduce the relations
$$\begin{align*} \l v_k' &= \Re \l \v_k \\ &= \Re(\g\v_i + \v_{k+1}) \\ &= \Re\bigl((a+bi)(v_k'+w_k'i)\bigr) + v_{k+1}' \\ &= av_k'-bw_k'+v_{k+1}'. \end{align*}$$
Similarly,
$$\l w_k' = bv_k'+aw_k'+w_{k+1}'.$$
Thus the matrix of $\l$ wrt the $v_k'$'s and $w_k'$'s has the block form
$$\begin{pmatrix} a & -b & 1 & &&&&&& \\ b & a & & 1 &&&&&& \\ && a & -b & 1 &&&&& \\ && b & a & & 1 &&&& \\ &&&& \ddots &&&&& \\ &&&&&& a & -b & 1 & \\ &&&&&& b & a & & 1 \\ &&&&&&&& a & -b \\ &&&&&&&& b & a \end{pmatrix}$$
I'll call this a real Jordan block. Thus, any linear map on a real vector space can be represented by a diagonal block matrix, where each block is either a usual Jordan block or a real Jordan block.
I discovered a matrix form slightly different from the real Jordan block. This form is obtained by considering the basis
$$\{p(\l), \l p(\l), p(\l)^2, \l p(\l)^2, \ldots\}$$
of $\R[\l]/p(\l)^k$, where $p(\l)=\l^2-\a\l-\b$ is an irreducible quadratic. Since
$$\l^2 p(\l)^i = -\b\ p(\l)^i - \a\l p(\l)^i + p(\l)^{i+1},$$
the matrix of $\l$ on this $\l$-span, wrt this basis, has the form
$$\begin{pmatrix} 0 & 1 & 0 & 0 &&&&&& \\ -\b & -\a & 1 & 0 &&&&&& \\ && 0 & 1 & 0 & 0 &&&& \\ && -\b & -\a & 1 & 0 &&&& \\ &&&& \ddots &&&&& \\ &&&&&& 0 & 1 & 0 & 0 \\ &&&&&& -\b & -\a & 1 & 0 \\ &&&&&&&& 0 & 1 \\ &&&&&&&& -\b & -\a \end{pmatrix},$$
composed of the subblocks
$$\begin{pmatrix} 0 & 1 \\ -\b & -\a\end{pmatrix} \quad\text{and}\quad \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}.$$