For matrix $A \in \mathbb{C}^{n\times n},$ it is well-known that there exists an invertible $P$ so that $PA=JP,$ where $J$ is the Jordan canonical form of $A$. Hence we have $e^A=P^{-1}e^JP.$ For real parameter $t$, if we want to compute $e^{tA},$ we could first compute the matrix $J=\text{diag}\{J_1,\cdots,J_k\},$ where $$ J_i=\begin{pmatrix} \lambda_i & 1 & \\ & \lambda_i & 1 & \\ & & \ddots & \ddots & \\ & & & \lambda_i & 1 \\ & & & & \lambda_i \end{pmatrix} \in \mathbb{R}^{n_i \times n_i}, $$ Then $$ e^{tJ_i}=e^{\lambda_i }\begin{pmatrix} 1 & t & \frac{t^2}{2} & \cdots & \frac{t^{n_i-1}}{(n_i-1)!} \\ & 1 & t & \cdots & \frac{t^{n_i-2}}{(n_i-2)!}\\ & & \ddots & \ddots & \vdots & \\ & & & 1 & t \\ & & & & 1 \end{pmatrix}. $$ Therefore $e^{tA}=P^{-1}\text{diag}\{e^{tJ_1},\cdots,e^{tJ_k}\}P.$
Recently I have read a more abstract but general way in a paper to express $e^{tA}.$ With the notations as we stated above, the paper said that (without any explanation) $$ e^{tA}=\sum_{i=1}^k \sum_{l=0}^{n_k-1} e^{t\lambda_i}t^l Z_{il}, $$ where $Z_{il}(i=1,\cdots,k;l=1,\cdots,n_k)$ are certain linearly independent matrices. How to deduce the above expression?
Motivated by @RobertIsrael 's answer, we should find polynomials $P_{il}$ for $1\leq i \leq k, 0 \leq l \leq n_k-1$ so that
$$ \frac{d^p}{dx^p}e^{tx}|_{x=\lambda}=\sum_{i=1}^k \sum_{l=0}^{n_i-1} e^{t\lambda_i}t^l P_{il}(\lambda) $$
for $\lambda=\lambda_1,\cdots,\lambda_k,p=0,\cdots,n_i-1,i=1,\cdots,k.$
The difficult is $\lambda_1$ could be equal to $\lambda_2$ since the diagonal elements for two Jordan blocks can coincide. Let us consider a simple case, assume $k=2$, i.e there are two Jordan blocks, and $\lambda_1=\lambda_2.$ We also assume the order of the two blocks are $n_1=2$ and $n_2=2$ respectively.
Since $e^{t\lambda_1}t^l=\frac{d^l}{dx^l}e^{tx}|_{x=\lambda_1},$ we should choose the polynominal so that
$$P_{10}(\lambda_1)+P_{20}(\lambda_1)=1,$$
$$P_{11}(\lambda_1)+P_{21}(\lambda_1)=0,$$
$$P_{10}^{'}(\lambda_1)+P_{20}^{'}(\lambda_1)=0,$$
$$P_{11}^{'}(\lambda_1)+P_{21}^{'}(\lambda_1)=1.$$
When I found these polynomials, how to show that $$ P_{il}(A), \quad i=1,2, l=1,2 $$ are linearly independent?
Look up: Holomorphic functional calculus. More generally, if $f$ is any function analytic in a neighbourhood of $\sigma(A)$ (the set of eigenvalues of $A$), then $f(A) = p(A)$ where $p(z)$ is a polynomial of degree $\le n$ such that, for each eigenvalue $\lambda$ of $A$ with (algebraic) multiplicity $m(\lambda)$, $p^{(k)}(\lambda) = f^{(k)}(\lambda)$ for $0 \le k \le m(\lambda)-1$. This lets you write $$ f(A) = \sum_{\lambda \in \sigma(A)} \sum_{k=0}^{m(\lambda)} f^{(k)}(\lambda) Z_{\lambda,k}(A)$$ where $Z_{\lambda,k}(z)$ is a polynomial such that $Z_{\lambda,k}^{(k)}(\lambda) = 1$, $Z_{\lambda,k}^{(j)}(\lambda) = 0$ for all other $j \in \{0,\ldots,m(\lambda)\}$, and $Z_{\lambda,k}^{(j)}(\mu) = 0$ for $0 \le j \le m(\mu)$ where $\mu \ne \lambda$ is another eigenvalue of $A$.
[EDIT] Note that each $\lambda$ is listed once here. In your example, with with two Jordan blocks of order $2$ but only one $\lambda$, we want $Z_{\lambda,0}(\lambda) = 1$, $Z'_{\lambda,0}(\lambda) = 0$ and $Z_{\lambda,1}(\lambda) = 0$, $Z'_{\lambda,1}(\lambda) = 1$: we can take $Z_{\lambda_0}(z) = 1$ and $Z_{\lambda,1}(z) = z-\lambda$. The result is $$ f(A) = f(\lambda) I + f'(\lambda) (A-\lambda I)$$