I see that there have been many questions on the strong Markov property, including Strong Markov property - Durrett and Two definitions of the strong Markov property.
I am still slightly confused about one item:
From my understanding of the strong Markov property, if $T$ is a stopping time, $X_n$ is a Markov chain (say on a finite state space $\mathcal{S}$), and $\mu$ denotes the initial distribution of $X_0$ then Durrett says (for example) \begin{equation} \mathbb{E}_{\mu}(X_{T+1}|\mathcal{F}_T) = \mathbb{E}(X_{T+1}|X_T) \quad \text{on } \{T < \infty\}. \end{equation} My question is: On the left-hand side, if we are conditioning on the stopping time $\sigma$-algebra $\mathcal{F}_T$, we already 'know' all the information up to the stopping time $T$ (in particular, information about the value taken by $X_0$). So shouldn't in this situation \begin{equation} \mathbb{E}_{\mu}(X_{T+1}|\mathcal{F}_T) = \mathbb{E}(X_{T+1}|\mathcal{F}_T)? \end{equation} If so, I suppose that this equality cannot hold in general, otherwise authors would have not bothered to include the subscript $\mu$. Why? (If I can improve the question clarity please let me know.)
Look very carefully at equations (1) and (2) in this answer. The right hand side of (1) is not a conditional expectation, and does not depend on the initial distribution.
On the other hand, your equation \begin{equation} \mathbb{E}_{\mu}(X_{T+1}|\mathcal{F}_T) = \mathbb{E}(X_{T+1}|X_T) \quad \text{on } \{T < \infty\}. \tag{incorrect}\end{equation} has conditional expectations on both sides, and should be written \begin{equation} \mathbb{E}_{\mu}(X_{T+1}|\mathcal{F}_T) = \mathbb{E}_{\mu}(X_{T+1}|X_T) \quad \text{on } \{T < \infty\},\tag{correct} \end{equation} as you thought.