A homogeneous Markov chain $\{X_n\}_{n\in\mathbb N}$ with discrete state space $\mathcal{S}$.
Set $$\tau_{k}:=\text{inf} \left\{n\ge 0:\, X_n=k \right\}.$$ where $\tau_{k}$ is defined to be $+\infty$, when doesn't exist any $n$ such that $X_n=k$.
$\textbf{1}.$
Now consider the Ehrenfest model,its state space $\mathcal{S}$ is a finite set $\left\{0,1,2,...,N\right\}$ and one-step transition matrix is
$$\left(\begin{array}{ccccccc} 0 & 1 & & & & & \\ \frac{1}{N} & 0 & \frac{N-1}{N} & & & & \\ & \frac{2}{N} & 0 & \ddots & & & \\ & & \frac{3}{N} & \ddots & \ddots & & \\ & & & \ddots & 0 & \frac{2}{N} & \\ & & & & \frac{N-1}{N} & 0 & \frac{1}{N} \\ & & & & & 1 & 0 \end{array}\right).$$
From an intuitive perspective, let $T_{ij}:=\inf\left\{n\ge 0:X_{n}=j,X_{0}=i\right\}$,$i<j$,be the hitting time of reaching $j$ starting from $i$.Since $T_{i,N}=\sum_{k=i}^{N-1}T_{k,k+1},$
then $${\color{Red} {\text{ for any } i<N,\mathbb{E}(\tau_{N}\mid X_{0}=i)=\mathbb{E}(\tau_{i+1}\mid X_{0}=i)+\mathbb{E}(\tau_{i+2}\mid X_{0}=i+1)+\cdots+\mathbb{E}(\tau_{N}\mid X_0=N-1).（*）}}\quad$$ But how can it be proven rigorously?
By linearity of expectation,we have
$$\begin{align} \mathbb{E}\left[\tau_{N}\mid X_{0}=i\right]&=\mathbb{E}\left[\tau_{i+1}+\sum_{t=i}^{N-2}(\tau_{t+2}-\tau_{t+1})\mid X_{0}=i\right] \\ &=\mathbb{E}\left(\tau_{i+1}\mid X_{0}=i\right)+\mathbb{E}\left(\tau_{i+2}-\tau_{i+1}\mid X_{0}=i\right)+\\&\cdots+\mathbb{E}\left(\tau_{N}-\tau_{N-1}\mid X_0=i\right). \end{align}$$
How to prove that formally
\begin{align} &\mathbb{E}\left(\tau_{i+2}-\tau_{i+1}\mid X_{0}=i\right)=\mathbb{E}\left(\tau_{i+2}\mid X_{0}=i+1\right);\\ &\qquad\qquad\qquad\qquad\cdots\cdots \\ &\mathbb{E}\left(\tau_{N}-\tau_{N-1}\mid X_0=i\right)=\mathbb{E}\left(\tau_{N}\mid X_0=N-1\right) .\end{align}
$\textbf{2}.$
Does this conclusion $(*)$ still hold in general homogeneous Markov chain $\{X_n\}_{n\in\mathbb N}$ with discrete state space $\mathcal{S}$?
A modest attempt on $\textbf{2}$：
Let the state space $\mathcal{S}$ is $\left\{0,1,2\right\}$ and one-step transition matrix is
$$ \begin{pmatrix} 1/3& 1/3& 1/3\\ 0& 1/2& 1/2\\ 0& 0& 1\\ \end{pmatrix}$$
I find that $$\mathbb{E}\left(\tau_{2}\mid X_{0}=0\right){\color{Red} \ne }\mathbb{E}\left(\tau_{1}\mid X_{0}=0\right)+\mathbb{E}\left(\tau_{2}\mid X_{0}=1\right).$$
Any help will be greatly appreciated！
This is an application of the strong Markov property. I will be assuming some familiarity with stopping times and Markov chains; you can refer to J.F. Le Gall's lecture notes "Measure Theory, Probability, and Stochastic Processes", more precisely chapters 12.2 and 13 and particularly Theorem 13.8.
Here the $\tau_i$ are stopping times. Indeed, writing $\mathcal F_n = \sigma(X_0, \dots, X_n)$ for the canonical filtration, for every $n$ the event $\{\tau_i \leq n\} = \cup_{k\leq n} \{X_k = i\} \in \mathcal F_n$. I also write $\mathbb E_j = \mathbb E[\, \dots \ | \ X_0=j]$.
In addition, for every $j>i$, on the event $\{X_0 = i\}$ we have $\tau_{j+1}-\tau_j = \tau_{j+1} \circ \Theta_{\tau_j} $, where $\Theta$ is the shift operator. This has to be proved of course. It is not too hard to understand why if you plot the trajectories: the shift operator $\Theta_t$ makes the trajectory start from time $t$ instead of time $0$, and if you start at time $\tau_j$ (thus at $X_0=j$), the number of steps before reaching $j+1$ (which is $\tau_{j+1}$ in the new trajectory) equals $\tau_{j+1}-\tau_j$ in the non-translated trajectory. The formal proof follows this reasoning almost exactly.
The strong Markov property (Theorem 13.8) then ensures that
$$ \mathbb E_i[\mathbf{1}_{\tau_{j} < \infty} (\tau_{j+1}-\tau_{j})] = \mathbb E_i[\mathbf{1}_{\tau_j < \infty} \tau_{j+1}\circ \Theta_{\tau_j}] = \mathbb E_i[\mathbf{1}_{\tau_j < \infty} \mathbb E_j[\tau_{j+1}]] = \mathbb E_j[\tau_{j+1}] \mathbb P_i(\tau_j < \infty) . $$ One only has to check that $\mathbb P_i(\tau_j < \infty)=1$ to get what you required. This comes from irreducibility and finiteness of the Markov chain.
For your second question, the result of 1. only holds under the specific condition that the only way to reach state $i+1$ from some state $j<i$ is to visit $i$ first. The spot where the proof fails is in the equality $\tau_{j+1}-\tau_j = \tau_{j+1}\circ \Theta_{\tau_j}$.