Now In (2.8) I have a really hard time understanding why we integrate with respect to $x_2$.
$P\left(x_3,t_3|x_1,t_1\right)=\sum _n P\left(x_3,t_3|x_n,t_2\right)P\left(x_n,t_2|x_1,t_1\right)$
This to me in a discrete case looks correct. But integrating over $x_2$ to me from the picture looks really strange. Is by integrating over $x_2$ am I integrating over the entire vertical line at $t_2$ and in that case it is in the x direction hence why not why not integrate over dx instead of $dx_2$
And most important which variable would you integrate on if there where more steps. Example $p(x_5,t_5|x_1,t_1)$
As it is said in the comments: $x_1$ at time $t_1$ and $x_3$ at time $t_3$ are fixed values. Next you fix $t_2 \in [t_1, t_3]$ and look at all possible values the process can attain at time $t_2$. For convenience this value is often called $x_2$ but you also can call it $x$, if you prefer.
However $x_2$ is not a fixed value (contrary to $x_1$ and $x_3$). So the picture might be misleading: there it seems like $x_2$ is also fixed and lies between $x_1$ and $x_3$, but this is just one (of many) possibilities. And now you sum up (or integrate) over all values $x_2$ the process can reach at time $t_2$, that's how you arrive at $$ P(x_3, t_3\mid x_1, t_1) = \sum_{x_2} P(x_2, t_2 \mid x_1, t_1) P(x_3, t_3 \mid x_2, t_2). $$ Note that the right hand side does not change if you write $x$ instead of $x_2$ $$ P(x_3, t_3\mid x_1, t_1) = \sum_{x} P(x, t_2 \mid x_1, t_1) P(x_3, t_3 \mid x, t_2). $$ It is just for convenience that the state at time $t_2$ is labeled $x_2$.