Writing $\exp(t \cdot H)$ where $H$ is a matrix and $t$ is a variable.
$\exp(t \cdot H) = \exp(t \cdot H_{0} + H_{1}) \cdot P_{0}$, $P_{0}$ represents the initial state
Create a generating function:

$$ \sum_{k=0}^{\infty} s_{k}z^{k} = \sum_{k=0}^{\infty} \dfrac{z^{k}}{k!} \partial^{k}{z}\left[ \exp(t(H{0} + H_{1}z)) \right ]{z=0}P{0} $$

Note that this involves using multiple derivatives which you can treat as recursive derivative. Look at the following iterated derivative

Found at

Mathlib.Analysis.Calculus.ContDiff.Defs

noncomputable def iteratedFDeriv (n : ℕ) (f : E → F) : E → E[×n]→L[𝕜] F :=
  Nat.recOn n (fun x => ContinuousMultilinearMap.curry0 𝕜 E (f x)) fun _ rec x =>
    ContinuousLinearMap.uncurryLeft (fderiv 𝕜 rec x)
#align iterated_fderiv iteratedFDeriv

/-- Formal Taylor series associated to a function within a set. -/
def ftaylorSeries (f : E → F) (x : E) : FormalMultilinearSeries 𝕜 E F := fun n =>
  iteratedFDeriv 𝕜 n f x
#align ftaylor_series ftaylorSeries

Use The Definition of Matrix Exponential:

$$ \sum_{k=0}^{\infty} \dfrac{z^{k}}{k!} \partial^{k}{z}\left[ \exp(t(H{0} + H_{1}z)) \right ]{z=0}P{0} = \sum_{k=0}^{\infty} \dfrac{z^{k}}{k!} \partial^{k}{z}\left[ \sum^{\infty}{l=0}\dfrac{(t(H_{0} + H_{1}z))^{l}}{l!} \right ]{z=0}P{0} $$

Now consider the following:

$$ \sum_{k=0}^{\infty} \dfrac{z^{k}}{k!} \partial^{k}{z}\left[ \sum^{\infty}{l=0}\dfrac{(t(H_{0} + H_{1}z))^{l}}{l!} \right ]{z=0} \cdot P{0}

= \sum_{k=0}^{\infty} \dfrac{z^{k}}{k!} \partial^{k}{z}\left[ \sum^{\infty}{l=0}\left (\dfrac{(\sum_{j = 0}^{1}((tH_{0})^{1-j}(H_{1}zt)^{j})))^{l}}{l!}\right) \right ]{z=0}P{0}

Here are we did was turn our addition into a summation sign using a transformation where we introduce an index j. Note that

here you can see if $j=0$ then $tH_{0}$ is picked and the latter is picked when $j=1$.

We will now move the $l!$ outside and turn our exponentiation to a product.

$$ \sum_{k=0}^{\infty} \dfrac{z^{k}}{k!} \partial^{k}{z}\left[ \sum^{\infty}{l=0}\left (\dfrac{(\sum_{j = 0}^{1}((tH_{0})^{1-j}(H_{1}zt)^{j})))^{l}}{l!}\right) \right ]{z=0}P{0}

\sum_{k=0}^{\infty} \dfrac{z^{k}}{k!} \partial^{k}{z}\left[ \sum^{\infty}{l=0} \dfrac{1}{l!}\prod_{i=0}^{l}\left(\sum_{j = 0}^{1}((tH_{0})^{1-j}(H_{1}zt)^{j}))\right) \right ]{z=0}P{0} $$

Furthermore lets sub-index the $j$ that we see with the summation.