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 $b=1$.
Best Answer
The relation $UT=TU=I$ is false, since you don't know whether $U$ is the inverse of $T$.
If $v\in E_\lambda$, then $Tv=\lambda v$ and $$ T(U(v)) = U(T(v)) = \lambda U(v) $$ so $U(v)\in E_\lambda$. This is enough to say $U(E_\lambda)\subseteq E_\lambda$ and since $U$ is invertible, it must necessarily hold $U(E_\lambda)=E_\lambda$.
There exists an index $m$ for which $K_\lambda = N((T-\lambda I)^m)$. Notice that $$ UT^k = T^kU \quad\forall k\implies U(T-\lambda I)^m = (T-\lambda I)^mU $$ so if $v\in K_\lambda$, then $$ 0 = U(T-\lambda I)^m(v) = (T-\lambda I)^m(U(v)) $$ and $U(v)\in N((T-\lambda I)^m)=K_\lambda$, so $U(K_\lambda)\subseteq K_\lambda$ and since $U$ is invertible, it must necessarily hold $U(K_\lambda)=K_\lambda$.