Here is a fundamental example (if $G$ is finite solvable, then $G/\Phi(G)$ is of the form described here).
Let $H$ be a group of invertible $n \times n$ matrices over the field $\mathbb{Z}/p\mathbb{Z}$. Let $N$ be the elementary abelian group of order $p^n$, $N \cong C_p^n$. Then $H$ acts on $N$ by matrix multiplication. Let $G = N \rtimes H$.
Then $G' = [G,G] = [N,N][N,H][H,H] = [N,H] \rtimes H'$.
$[N,H] = \langle n^{-1} n^h : n \in N, h \in H \rangle$ in multiplicative notation, but in matrix notation, we just get $$[N,H] = \langle -n + n \cdot h : n \in N, h \in H \rangle = \langle n\cdot (h-1): n \in N, h \in H \rangle = \sum_{h \in H} \newcommand{\im}{\operatorname{im}}\im(h-1)$$
Coprime action
If $H$ has order coprime to $p$, then some fancy linear algebra shows that $\ker((h-1)^n) = \ker(h-1)$ and $\im(h-1)$ is a direct complement. In other words, by Fitting's lemma (applied to the semisimple operator $h-1$), we get $$N = \ker(h-1) \oplus \im(h-1) = C_{h}(N) \times [N,h]$$
Using some slightly fancier versions of these linear algebra ideas we even get $$N=\left( \bigcap_{h \in H} \ker(h-1) \right) \oplus \left(\sum_{h \in H} \im(h-1) \right) = C_H(N) \times [H,N]$$
Even if $N$ is not abelian similar ideas give: $N=C_H(N)[H,N]$, though the intersection may be non-identity.
Defining characteristic
If $H$ has order a power of $p$, then one gets sort of the opposite behavior. The minimum polynomial of $h$ divides $x^{p^n}-1 = (x-1)^{p^n}$, so every eigenvalue of $h-1$ is 0, and $h-1$ is nilpotent. Hence Fitting's lemma tells us that $N=\ker((h-1)^{p^n}) \times \im((h-1)^{p^n})$, but that is useless since $(h-1)^{p^n}=0$ and so the kernel is all of $N$ and the image is $1$.
If we try to apply this to $h-1$ directly without raising to the $p^n$th power, then things go very weird. Take $h=\begin{bmatrix}1&1\\0&1\end{bmatrix}$. Then $\im(h-1) = \{ (0,x) : x \in C_p \}$ but also $\ker(h-1) = \{ (0,x) :x \in C_p \}$. When $p=2$, this is the $D_8$ example.
If one wants larger $N$, then one can take $H=\langle h_1,h_2\rangle$ with $$
h_1=\begin{bmatrix}1&0&1\\0&1&0\\0&0&1\end{bmatrix}, \qquad
h_2=\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}
$$
Then $\im(h_i-1)=\{ (0,0,x) :x \in C_p \}$ but $\ker(h_1-1) = \{ (0,y,z) : y,z \in C_p \}$ and $\ker(h_2-1) = \{ (x,0,z) : x,z \in C_p \}$ so $$\bigcap_{h \in H} \ker(h-1) = \{ (0,0,x) : x \in C_p \}$$ and $$\sum_{h \in H} \im(h-1) = \{ (0,0,x) : x \in C_p \}$$
When $p=2$, this is the $D_8 \operatorname{\sf Y} D_8$ example.
Notice how broken the decomposition is here.
References
Kurzweil–Stellmacher, Theory of Finite Groups, Chapter 8, is where this really started to make sense for me.
