The cyclic decomposition theorem actually says that a cyclic decomposition always exists, since you can take $W_0=\{0\}$. Mentioning $W_0$ in that version of the cyclic decomposition theorem is just done in order to allow a proof by induction on the (remaining) dimension: by constructing an appropriate new cyclic factor so that its sum with the old $W_0$ is still $T$-admissible. (That $W_0$ is required to be a proper subspace is to avoid the case $r=0$, but it would have been better to just allow that case.)
Cyclic decompositions lead to "rational forms"; special matrices similar to the original one over the the original field (no field extension). These forms exist whether or not the eigenvalues (characteristic values) live over the base field, since blocks of the matrix are companion matrices of the annihilators of the cyclic factors which could be any polynomials, not just those of the form $X-\lambda$. Cyclic decompositions are not unique however, not even up to some coarse notion of equivalence; for instance the number of cyclic factors can vary among decompositions. There is a notion of Rational Canonical Form, in which the additional condition is imposed that the annihilators of the cyclic factors each divide the next (or in another flavour each are a multiple of the next); this does not yet make the cyclic decomposition unique, but it does make the sequence of annihilators unique, and therefore the associated rational form.
The rational canonical form corresponds to a decomposition in as few cyclic factors as possible. For instance if the whole space is cyclic it will just be a single companion matrix; moreover this is often the case, for instance if $T$ is diagonalisable without repeated eigenvalues, or more generally if the minimal polynomial equals the characteristic one. In general its decomposition contains one cyclic direct factor whose annihilator is as large as possible, namely equal to the minimal polynomial, while a complementary $T$-stable subspace is (if nonzero) similarly decomposed recursively (although this is not the method by which the decomposition is found). Since no decomposition of cyclic factors is attempted corresponding to factoring of minimal polynomials, the rational canonical form does not change when extending scalars to a larger field. As a consequence the entries of the rational canonical form live in the smallest field for which any matrix of $T$ can be found.
There is however a different decomposition that does reflect factoring of the minimal polynomial over the field at hand, which is the primary decomposition. It is not necessarily a decomposition into cyclic factors, but has the advantage over decompositions into cyclic factors that the decomposition itself is canonical: it does not depend on any choices (but it does depend on the field). As a consequence the primary decomposition is compatible with any $T$-invariant subspace $W\subseteq V$: it is always true that $W$ is direct sum of subspaces of the primary factors of$~V$, namely of its intersections with those factors, which intersections form the primary factors of$~W$. (Nothing of this kind holds for decompositions into cyclic factors corresponding to the rational canonical form).
A decomposition into a maximal number of cyclic factors (which are therefore as small as possible) can be obtained be decomposing each primary factor of$~V$ into cyclic factors (this decomposition is not unique, thought in this case all possible decompositions are isomorphic). The result is called the primary rational canonical form. The number of its cyclic factors may increase as one extends scalars to a larger field (if this leads to the minimal polynomial factoring into smaller irreducible factors).
This answer turned out far longer than initially planed:
It explains the connection between the Jordan-Chevalley decomposition and the Jordan normal form, why Petersen only considers a single Jordan block, and what the Jordan-Chevalley decomposition is useful for.
The connection between the Jordan-Chevalley decomposition and the Jordan normal form:
As it has already been explained in the comments, the Jordan-Chevalley decomposition of $T$ can be derived from its Jordan canonical form:
Suppose that $\mathcal{B}$ is a basis of $V$ with respect to which the operator $T$ is given by a matrix $[T] \in \operatorname{M}_n(\mathbb{C})$ which is in Jordan normal form, say
$$
[T]
= \begin{pmatrix}
J_{n_1}(\lambda_1) & & \\
& \ddots & \\
& & J_{n_t}(\lambda_t)
\end{pmatrix}.
$$
(Here the $\lambda_i$ are not necessarily pairwise distinct.)
Then with respect to $\mathcal{B}$ the operators $S$ and $N$ are given by the matrices
$$
\begin{pmatrix}
\lambda_1 I_{n_1} & & \\
& \ddots & \\
& & \lambda_t I_{n_t}
\end{pmatrix}
\quad\text{and}\quad
\begin{pmatrix}
J_{n_1}(0) & & \\
& \ddots & \\
& & J_{n_t}(0)
\end{pmatrix}.
$$
One could also go the other way around, and derive the Jordan normal form of $T$ from its Jordan-Chevalley decomposition:
Every eigenspace $V_\lambda(S)$ is $N$-invariant, since $S$ and $N$ commute.
Since $N$ is nilpotent, the same goes for the restrictions $N|_{V_\lambda(S)}$.
Thus we can find for every $\lambda \in \mathbb{C}$ a basis $\mathcal{B}_\lambda$ for $V_\lambda(S)$ with respect to which the operator $N|_{V_\lambda(S)}$ is given by a matrix which is in Jordan normal form, say $[N|_{V_\lambda(S)}] = \bigoplus_{j=1}^{n(\lambda)} J_{n(\lambda,j)}(0)$ (here we use that finite-dimensional nilpotent operators always have a Jordan normal form, and that $0$ is the only eigenvalue of a nilpotent operator).
Since $S$ is diagonalizable we have that $V = V_{\lambda_1}(S) \oplus \dotsb \oplus V_{\lambda_r}(S)$ (with the $\lambda_i$ being pairwise distinct), so it follows that the union $\mathcal{B} := \bigcup_{i=1}^r \mathcal{B}_{\lambda_i}$ is a basis of $V$.
With respect to $\mathcal{B}$ the operator $N$ is given by the block diagonal matrix $[N] = \bigoplus_{i=1}^r \bigoplus_{j=1}^{n(\lambda_i)} J_{n(\lambda_i, j)}(0)$, which is again in Jordan normal form, and the operator $S$ is given by the diagonal matrix $[S] = \bigoplus_{i=1}^r \lambda I_{\dim V_{\lambda_i}(S)}$.
So with respect to $\mathcal{B}$ the operator $T = S + N$ is given by the matrix
\begin{align*}
[T]
= [S] + [N]
&= \left( \bigoplus_{i=1}^r \bigoplus_{j=1}^{n(\lambda_i)} J_{n(\lambda_i, j)}(0) \right)
+ \left( \bigoplus_{i=1}^r \lambda I_{\dim V_{\lambda_i}(S)} \right)
\\
&= \bigoplus_{i=1}^r \bigoplus_{j=1}^{n(\lambda_i)} J_{n(\lambda_i, j)}(\lambda_i),
\end{align*}
which is in Jordan normal form.
Alltogether this shows that the Jordan-Chevalley decomposition and the Jordan normal form are equivalent, and how one can be derived from the other.
This observation actually holds for arbitary fields:
An operator $T \colon V \to V$ on a finite-dimensional $k$-vector space $V$ has a Jordan-Chevalley decomposition (into commuting diagonalizable and nilpotent parts) if and only if it has a Jordan normal form.
Also note that the decomposition
$$
V = \ker(T - \lambda_1 I)^{m_1} \oplus \dotsb \oplus \ker(T - \lambda_k I)^{m_k}.
$$
which is used to construct the Jordan-Chevalley decomposition is precisely the generalized eigenspace decomposition, which is used to show the existence of the Jordan normal form.
Regarding the number of Jordan blocks:
I am not very familiar with the Frobenius canonical form which Petersen uses here, but I think I (kind of) understand where the problem comes from, and how to solve it.
You are right that we may need more than one Jordan block if we look at the restriction of $T$ to $\ker (T - \lambda_i)^{m_i}$;
the matrix representation of $[T|_{\ker (T - \lambda_i)^{m_i}}]$ consists of all Jordan blocks for the eigenvalues $\lambda$.
This is why Petersen further decomposes $\ker (T - \lambda_i)^{m_i}$ into cyclic subspace:
This means that we have reduced the problem to a situation where $T$ has only one eigenvalues.
Given the Frobenius canonical form the problem is then further reduced to [proving] the statement for companion matrices, where the minimal polynomial has only one root.
Let $C_p$ be a companion matrix with $p(t) = (t - \lambda)^n$.
(From Linear Algebra by Peter Petersen, page 150, proof of Theorem 25.)
So we further decompose
$$
\ker (T - \lambda_i)^{m_i}
= C_1 \oplus \dotsb \oplus C_{k(i)}
$$
where the $C_j$ are cyclic subspaces.
We fix some $j$ and set $C := C_j$ and $n := \dim C$.
Since $C$ is cyclic, we find that the characteristic polynomal and minimal polynomial of $T|_C$ coincide (I assume that this has already been shown before); we will refer to this polynomial as $p$.
We know that this minimal polynomial $p(t)$ of $T|_C$ divides the minimal polynomial of $T|_{\ker (T - \lambda_i)^{m_i}}$, which is given by $(t - \lambda_i)^{m_i}$.
So $p(t)$ is of the form $p(t) = (t - \lambda_i)^{m'_i}$ with $m'_i \leq m_i$.
Together with $\deg p = \dim C = n$ we find that $p(t) = (t - \lambda_i)^n$.
We now consider the matrix
$$
J
:= \begin{pmatrix}
\lambda & 1 & & \\
& \ddots & \ddots & \\
& & \ddots & 1 \\
& & & \lambda
\end{pmatrix}
\in \operatorname{M}_n(\mathbb{C}).
$$
From an earlier part of the chapter (namely part 4, The Minimal Polynomial, page 120, Proposition 17) we know that the minimal polynomial of $J$ is given by $(t - \lambda)^n = p(t)$.
Since the minimal polynomial of $J$ is of maximal degree it equals its characteristic polynomial;
from this is follows that $J$ is similar to the companion matrix its characteristic polynomial $p(t)$, which we will refer to as $C_p$.
(Petersen seems to have shown this before, but gives no explicit reference in the proof.)
Since the minimal and characteristic polynomial of $T$ coincide, we find that when we represent $T$ with respect to some basis of $C$ by a matrix $A \in \operatorname{M}_n(\mathbb{C})$, then $A$ is also similar to the companion matrix $C_p$.
Hence we find that $A$ and $J$ and similar, so there exists a basis of $C$ with respect to which $T|_C$ is represented by $J$.
(There might be some redundancy in the above argumentation.)
Note that we have shown that the decomposition $\ker (T - \lambda_i)^{m_i} = C_1 \oplus \dotsb \oplus C_k$ into cylic subspaces corresponds precisely to the decomposition of $[T|_{\ker (T - \lambda_i)^{m_i}}]$ into Jordan blocks.
Since we restrict our attention to single cyclic subspace, we also get only Jordan block.
I have to admit that I find Petersen’s proof somewhat strange:
What he actually does is to construct the Jordan normal form by construction a decomposing $V = \bigoplus_{i=1}^k \bigoplus_{j=1}^{k'(i)} C_{\lambda_i, j}$ into cyclic subspaces $C_{\lambda_i, j}$, and then showing that for each $C_{\lambda_i, j}$ there exists a basis with respect to which $T|_{C_{\lambda_i, j}}$ is given by a matrix $[T|_{C_{\lambda_i, j}}]$ which is a Jordan block.
Then he constructs the Jordan-Chevalley decomposition from the Jordan normal form — without ever mentioning the Jordan normal form.
I suppose that this doesn’t help understanding the difference between the two constructions.
Advantages of the Jordan-Chevalley decomposition:
One way to think about the Jordan-Chevalley decomposition is to regard it as a coordinate-free version of the Jordan normal form:
To talk about the Jordan normal form of $T$ we need to associate to $T$ a matrix $[T]$, which requires the use of a basis.
The Jordan-Chevalley decomposition on the other hand has no such requirements.
What has not been mentioned so far, but is very useful, is that $S$ and $N$ can be expressed as polynomials of $T$, i.e. there exists polynomials $p(t), q(t) \in \mathbb{C}[t]$ with $S = p(T)$ and $N = q(T)$.
As far as I know, this has no analogue in terms of the Jordan normal form.
The Jordan-Chevalley decomposition also has the advantage that it generalizes more easily to other settings:
One can generalize the notion of a diagonalizable operator to that of a semisimple operator (if we work over an algebraically closed field then both notions coincide).
Then one can also generalize the Jordan-Chevalley decomposition accordingly.
One can generalize the Jordan-Chevalley decomposition to finite-dimensional, semisimple complex Lie algebras:
If $\mathfrak{g}$ is such a Lie algebra, then every elemente $x \in \mathfrak{g}$ can be uniquely written as $x = s + n$ where $s, n \in \mathfrak{g}$ are semisimple, resp. nilpotent elements which commute.
One can generalize the additive Jordan-Chevalley decomposition, which we have encountered so far, to the multiplicative Jordan-Chevalley decomposition: Every $T \in \operatorname{GL}_n(\mathbb{C})$ can be a uniquely decomposition as $T = S'U'$ with $S' \in \operatorname{GL}_n(\mathbb{C})$ being diagonalizable and $U' \in \operatorname{GL}_n(\mathbb{C})$ being unipotent.
(The additive and multiplicative Jordan-Chevalley decompositions $T = S + N$ and $T = S' U'$ are related by $S = S'$ and $U' = 1 + S^{-1} N$.)
Best Answer
Let $p$ be the minimal polynomial of $A$. Then there is a unique polynomial $s\in K[X]$ of degree less than $\deg p$ such that $s(A)$ is the semi-simple part of $A$.
Assume, as we may, that $p$ splits in $K$ as, say, $$ p=\prod_{\lambda\in\Lambda}\ (X-\lambda)^{m(\lambda)} $$ with $m(\lambda) > 0$ for all $\lambda$.
Then $s$ is the unique degree less than $\deg p$ solution to the congruences $$ s\equiv\lambda\quad\bmod\quad(X-\lambda)^{m(\lambda)},\quad\lambda\in\Lambda, $$ and it is given by $$ s=\sum_{\lambda\in\Lambda}\ \lambda\ T_\lambda\left(\frac{(X-\lambda)^{m(\lambda)}}{p}\right)\frac{p}{(X-\lambda)^{m(\lambda)}}\quad, $$ where $T_\lambda(f)$ means "order less than $m(\lambda)$ Taylor polynomial of $f$ at $\lambda$".
EDIT. Here is a proof. Put $$ B_\lambda:=\frac{K[X]}{(X-\lambda)^{m(\lambda)}}\quad. $$
(A) We have canonical $K[X]$-algebra isomorphisms $$ K[A]\simeq\frac{K[X]}{(p)}\simeq\prod_{\lambda\in\Lambda}\ B_\lambda=:B, $$ the second isomorphism being given by the Chinese Remainder Theorem.
We way (and will) work in $B$ instead of working in $K[A]$.
Let $x\in B$ be the canonical image of $X$, and $e_\lambda$ the element of $B$ whose $\lambda$ component is $1$, and whose other components are $0$.
We must find the semi-simple part of $x$. But this is clearly the sum of the $\lambda e_\lambda$. In view of (A), this shows that, as claimed, $s$ is the unique degree less than $\deg p$ solution to the congruences $$ s\equiv\lambda\quad\bmod\quad(X-\lambda)^{m(\lambda)},\quad\lambda\in\Lambda, $$ and we're left with solving these congruences.
It's not harder to solve the general congruence system $$ s\equiv p_\lambda\quad\bmod\quad(X-\lambda)^{m(\lambda)},\quad\lambda\in\Lambda, $$ where the $p_\lambda\in K[X]$ are arbitrary.
The trick is to use the Ansatz $$ s:=\sum_{\lambda\in\Lambda}\ s_\lambda\ \frac{p}{(X-\lambda)^{m(\lambda)}}\quad,\quad\deg s_\lambda < m(\lambda), $$ which gives the solution $$ \sum_{\lambda\in\Lambda}\ T_\lambda\left(p_\lambda\ \frac{(X-\lambda)^{m(\lambda)}}{p}\right)\frac{p}{(X-\lambda)^{m(\lambda)}}\quad. $$
[Recall that $A$ admits a Jordan decomposition if and only if its eigenvalues are separable over $K$ (Bourbaki, Algèbre, Théorème VII.5.9.1). We assume here that such is the case.]