Let $X$ be a jump process on a (for simplicity) finite state space $S$. Denote with $\mathcal{F}^X$ the filtration generated by $X$, so that $\mathcal{F}_t^X$ is all the information generated by observing $X$ up until and including time $T$. Let $T(t)$ be the next jump time of $X$ after time $t$, i.e. $T(t) = \inf\{s \in (t,\infty) : X(s) \neq X(t)\}$.
To understand the differences between Markov and semi-Markov processes, the following suffices:
If $X$ is time-homogeneous Markov with transition rates $\mu_{jk}$, $j,k \in S, k \neq j$, then $T(t)-t$ given $\mathcal{F}^X_t$ is exponentially distributed with rate $\mu_{X(t)\bullet} := \sum_{k \neq X(t)} \mu_{X(t)k}$.
If $X$ is time-inhomogeneous Markov with transition rates $\mu_{jk}(t)$, now time-dependent, then $T(t)-t$ given $\mathcal{F}^X_t$ has density $s \mapsto \mu_{X(t)\bullet}(s) e^{-\int_t^s \mu_{X(t)\bullet}(u) \, \mathrm{d}u}$. (If the transition rates are time-independent, then this is the density of an exponential distribution, so we have a consistent extension of the time-homogeneous case.)
In the (time-homogeneous) semi-Markov case, we need the duration process $U(t) := t- \sup\{s \in [0,t] : Z(s) \neq Z(t)\}$, which is the time since the last jump of $X$. If $X$ is time-inhomogeneous Markov with transition rates $\mu_{jk}(u)$, then $T(t)-t$ given $\mathcal{F}_t^X$ has density $s \mapsto \mu_{X(t)\bullet}(U(t-)+s) e^{-\int_t^{t+s}\mu_{X(t)\bullet}(U(t-)+u-t) \, \mathrm{d}u}$, which depends on the current state $X_t$ as well as the current duration.
The time-inhomogeneous and semi-Markov case might look similar, but they are different. Consider the time interval $[0,2]$. Consider two cases. In both cases, we only have one jump. In the first case, the jump occurs at time $1$, and no further jumps occur up until time $2$. In the second case, a jump occurs at time $2$. We are interested in the distribution of the next jump time $T(2)$ minus 2. If $X$ is time-inhomogeneous Markov, we must in both cases consider the same density $\mu_{X(2)\bullet}(s) e^{-\int_2^s \mu_{X(2)\bullet}(u) \, \mathrm{d}u}$. If $X$ is semi-Markovian, the density in the first case is $\mu_{X(2)\bullet}(1+s) e^{-\int_0^s \mu_{X(2)\bullet}(1+u) \, \mathrm{d}u}$, while it is $\mu_{X(2)\bullet}(s) e^{-\int_0^s \mu_{X(2)\bullet}(u) \, \mathrm{d}u}$ in the second case.
Really, there are two 'clocks' in play. There is the 'time' clock, which ticks continuously throughout time, and then there is the 'duration' clock, which is reset at every jump. If you imagine that $X$ are health events of an individual, what is going one can be explained as follows. If $X$ is time-inhomogeneous Markov, the time until the next health event (healthy, ill, dead) is allowed to depend on the current health state of the individual (health, ill, dead) and his age (time). If $X$ is (time-homogeneous) semi-Markov, the time until the next health event is allowed to depend on the current health state of the individual and the time since he entered this state of health (duration). These two characteristics are of course very different.
Both aspects are combined in the time-inhomogeneous semi-Markov case with transition rates $\mu_{jk}(t,u)$ (that depend on time as well as duration). Here the time until the next health event is allowed to depend on the current health state of the individual, the time since he entered this state of health, and his sage.
Best Answer
Yes, $x$ is time spend on a state, and $\tau_n$ is the time of the $n$-th jump. To clarify this, define $T_n=\tau_n-\tau_{n-1}$ for $n\geq 1$ and $\tau_0=0$. In this sense, $T_n$ is the time between jumps and $\tau_n=T_1+\cdots +T_n$.
To understand $F_{ij}(x)$ use the above definition of $T_n$ and define $X_n=X(\tau_n)$. Manipulating the probability presented in the reference you get that \begin{align} \mathbb{P}(T_n\leq x,X_n=j|X_{n-1}=i)&=\dfrac{\mathbb{P}(T_n\leq x,X_n=j,X_{n-1}=i)}{\mathbb{P}(X_{n-1}=i)}\\ &=\dfrac{\mathbb{P}(T_n\leq x|X_n=j,X_{n-1}=i)\mathbb{P}(X_n=j,X_{n-1}=i)}{\mathbb{P}(X_{n-1}=i)}\\ &=\mathbb{P}(T_n\leq x|X_n=j,X_{n-1}=i)\mathbb{P}(X_n=j|X_{n-1}=i)\\ &=\mathbb{P}(T_n\leq x|X_n=j,X_{n-1}=i)p_{ij}. \end{align} So $F_{ij}(x):=\mathbb{P}(T_n\leq x|X_n=j,X_{n-1}=i)$. This is similar to the distribution of the observed process in a two-step hidden markov model where the regime is $\{X_n\}$.