What we will do is construct a generalization of the real Jordan canonical form. I will mostly omit proofs, since they are simple computations. This will work for any endomorphism $\phi$ over a finite dimensional $\mathbb{K}$-vector space $V$ such that the splitting field of the polynomial characteristic of $\phi$ is of degree $2$.
Let $L$ be said extension of $\mathbb{K}$, i.e. $L = \mathbb{K}(\alpha)$. Every element of $L$ can be written as $z = x + \alpha y$, with $x$ and $y \in \mathbb{K}$. In lack of a better notation, since i do not know if there is one, i will use $\text{Re}(z)$ to mean $x$ and $\text{Im}(z)$ for y, and call them respectively real and imaginary part of $z$, taking inspiration from the real and complex numbers. If you know a better one, or a more used one, please let me know.
We consider now the endomorphism $\phi \otimes Id_L$ of $V \otimes_\mathbb{K} L$, and by abuse of notation denote it by $\phi$ again. In the real case, this corresponds to complexification. The important thing is that the matrix representing $\phi$ and $\phi \otimes Id_L$ are the same, and have the same characteristic polynomial (since the matrices are the same) and the same minimal polynomial (see, for example Minimal polynomial is invariant under field extensions).
We first note that the spectrum of $\phi$ is the union of the eigenvalues belonging to $\mathbb{K}$ and those belonging to $L \setminus \mathbb{K}$. Those belonging to $L \setminus \mathbb{K}$ come in "conjugate" pairs, in the sense that if $\mu = a + \alpha b$ is an eigenvalue, $\bar{\mu} = a - \alpha b$ is another eigenvalue and both have the same algebraic multiplicity. Denoting $V(\lambda)'$ the generalized eigenspace relative to the eigenvalue $\lambda$, we can decompose $V \otimes L$ as $V \otimes L = \bigoplus_{\lambda_j} V(\lambda_j)'$. This is possible since the characteristic polynomial of $\phi$ is split in $L[t]$, so it admits a Jordan canonical form as $L$-matrix. For the eigenvalues belonging to $\mathbb{K}$, all $\ker(\phi - \lambda I)^j$ have the same dimensions both when seeing them as $\mathbb{K}$-vector spaces and as $L$-vector spaces, so we can find a real Jordan basis for $V(\lambda)$. Regarding the eigenvalues in $L \setminus \mathbb{K}$, it can be easily proven that if $\{z_1, \dots, z_k\}$ is a Jordan basis for $V(\lambda)'$, then $\{\bar{z_1}, \dots, \bar{z_k}\}$ is a Jordan basis for $V(\bar{\lambda})'$. Moreover, there is a basis made of elements of $V$ for $ V(\lambda)' \oplus V(\bar{\lambda})'$ by noting that $\text{Re}(z) = \frac{z + \bar{z}}{2}$ and $\text{Im}(z) = \frac{z - \bar{z}}{2 \alpha}$. From this, it is easy to see the real and imaginary parts of the elements of the basis $\{z_1, \dots..., z_k, \bar{z_1}, \dots, \bar{z_k}\}$ form a basis.
We can now find, supposing $\mu \in L \setminus \mathbb{K}$, what is the matrix associated to $\phi$ restricted to $V(\mu)' \oplus V(\bar{\mu})'$ with respect to a real basis of a Jordan basis. I will denote the real parts of $z$ by $x$ and the imaginary ones by $y$, and let $\{z_1, \dots, z_k, \bar{z_1}, \dots, \bar{z_k}\}$ a Jordan basis for $V(\mu)' \oplus V(\bar{\mu})$.
If $z_j$ is such that $\phi (z_j) = \mu z_j$, then $$\phi(x_j) = \phi(\frac{z_j + \bar{z_j}}{2}) = \text{Re}(\mu z_j) = \frac{\mu z_j + \bar{\mu} \bar{z_j}}{2} = \text{Re}(\mu)x_j + \alpha^2 \text{Im}(\mu)y_j$$ and $$\phi (y_j) = \text{Im}(\mu z_j) = \text{Im}(\mu)x_j + \text{Re}(\mu)y_j$$
If, on the other hand, $z_j$ is such that $\phi(z_j) = \mu z_j + z_{j-1}$, we get $$\phi(x_j) = \phi(\frac{z_j + \bar{z_j}}{2}) = \frac{\mu z_j +z_{j-1} + \bar{\mu} \bar{z_j} + \bar{z_{j_1}}}{2} = \text{Re}(\mu)x_j + \alpha^2 \text{Im}(\mu)y_j + x_{j-1}$$ and $$\phi(y_j) = \text{Im}(\mu)x_j + \text{Re}(\mu)y_j + y_{j-1}$$
Then it is immediate to find the matrix representing $\phi$ restricted to $\text{Span} \{x_i, y_i \}$ wrt the basis $\{x_i, y_i \}$:
$$
\begin{bmatrix}
\text{Re}(\mu) & \alpha^2 \text{Im}(\mu) \\
\text{Im}(\mu) & \text{Re}(\mu)
\end{bmatrix}
$$
In our case, the eigenvalue is $\mu = \sqrt{3} = 0 + 1 \sqrt{3}$ so the matrix will be
$$
\begin{bmatrix}
0 & 3 \\
1 & 0
\end{bmatrix}
$$
and the Jordan generalized form is
$$
\begin{bmatrix}
J(2, 2) & 0 & 0 & 0 & 0 \\
0 & J(2, 2) & 0 & 0 & 0 \\
0 & 0 & J(\sqrt{3}, 2) & 0 & 0 \\
0 & 0 & 0 & J(\sqrt{3}, 2) & I_2 \\
0 & 0 & 0 & 0 & J(\sqrt{3}, 2)
\end{bmatrix}
$$
where $J(2, 2)$ is a block of size two corresponding to the eigenvalue 2, and $J(\sqrt{3}, 2)$ is the $2 x 2$ matrix written above and $I_2$ is an identity block of size $2$.
Best Answer
When the characteristic polynomial does not split into linear factors I would approach this as follows.
A way to prove the existence of most of the canonical forms is the observation that having a linear transformation $T:V\to V$ allows us to turn $V$ into an $\Bbb{F}[x]$-module by letting $x$ act as $T$. More precisely, if $v\in V$ and $p(x)=\sum_{i=0}^na_ix^i\in\Bbb{F}[x]$ are arbitrary, we define $$ p(x)\cdot v=\sum_{i=0}^na_i T^i(v). $$ Anyway, let's see where this approach takes us.
As $V$ was assumed to be finite dimensional, it becomes a finitely generated $\Bbb{F}[x]$-module. A polynomial ring over a field is a PID (a Euclidean domain even), so the structure of f.g. modules over a PID allows us to write $V$ as a direct sum oof cyclic modules $$ V=\bigoplus_{j=1}^rV_j, $$ where each summand $V_j\simeq \Bbb{F}[x]/\langle p_j(x)\rangle$ for some non-constant polynomial $p_j(x)\in\Bbb{F}[x]$.
If $p_j(x)=\prod_\ell q_\ell(x)^{a_\ell}$ is a factorization of $p_j(x)$ into a product of powers of irreducible polynomials $q_\ell(x)$, then the polynomial variant of the Chinese Remainder Theorem allows us to further split the submodules $$ \Bbb{F}[x]/\langle p_j(x)\rangle=\bigoplus_\ell\Bbb{F}[x]/\langle q_\ell(x)^{a_\ell}\rangle. $$ Putting this all together we see that $V$ can be written as a direct sum of modules of the form $\Bbb{F}[x]/\langle r(x)^a\rangle$ for some irreducible polynomial $r(x)\in\Bbb{F}[x]$ and positive integer $a$. Such a decomposition (into submodules) is automatically also a decomposition into $T$-stable subspaces.
Let's extract one more thing from the theory of modules over a PID. To each irreducible polynomial $r(x)$ we can define the $r$-power torsion submodule $$ V_r:=\{v\in V\mid r(x)^k\cdot v=0\ \text{for some integer $k>0$}\}. $$ In my opinion the subspaces $V_r$ are the natural analogue of the generalized eigenspaces. After all, should $\Bbb{F}$ be algebraically closed, we would necessarily have $r(x)=x-\lambda$ for some $\lambda\in\Bbb{F}$, and the above definition of $V_r$ becomes the definition of a generalized eigenspace.
The submodule $V_r$ is the direct sum of all those earlier summands $\Bbb{F}[x]/\langle q_\ell(x)^{a_\ell}\rangle$ where the irreducible polynomial $q_\ell(x)=r(x)$.
My main point was that when the characteristic polynomial doesn't split into linear factors, it feels natural to lump together conjugate eigenvalues, i.e. the zeros of an irreducible factor of the characteristic polynomial. The above subspaces $V_r$ become, upon the appropriate extension of scalars, the sum of generalized eigenspaces belonging to all those conjugate eigenvalues.
It is easy to build an analogue of a Jordan block in this setting. A submodule of the form $\Bbb{F}[x]/\langle r(x)^a\rangle$ then corresponds with an $ab\times ab$ matrix, $b=\deg r(x)$. Along its diagonal we have $b\times b$-blocks representing $r(x)$. We can arrange those diagonal blocks to become, say, companion matrices of $r(x)$. Then, on top of the main diagonal we have blocks of the form $I_b$, analogous to those extra $1$ on top of the diagonal of a Jordan block in the case $b=1$.