I am reading Tutte's theorem proof:
(The part which I don't understand follows).
We have a graph $G$ with vertices $x,a,b,c\in V(G)$ s.t. $ac,xb\notin E(G)\ni xa,ab$. We are given that $G$ hasn't a perface matching but $G+xb$ has a perfact matching $M_1$ with $xb\in M_1$ and $G+ac$ has a perfact matching $M_2$ with $ac\in M_2$.
In the proof, a path $P$ is defined to be a maximal path that starts from $c$ such that $P$ is altermating between edges from $M_1$ and edges from $M_2$.
I don't understand why if $P$ ends up with an edge from $M_2$ then necesserily the last vertex in $P$ has to be $x$ or $b$.
Note that $P$ is a path in $G$. I.e., it is not allowed to use either of the added edges $xb$ or $ac$.
$P$ starts at $c$. It is not allowed to follow $ac$, which is the edge connecting to $c$ in $M_2$, so it has to instead use the edge for $c$ from $M_1$. For any vertex $v$ other than $a,b,c,x$, when it arrives at $v$ by the edge from $M_1$, it can leave by the edge from $M_2$, and if it arrives by the edge from $M_2$ it can leave by the edge from $M_1$. There is nothing to stop $P$ from continuing at any of these vertices.
But with $a, c, b$ or $x$, things are different. It cannot arrive at $a$ or $c$ by an edge in $M_2$, because the edge for these points is $ac$, which is not in $G$. So it arrives at $a$ by an edge in $M_1$, and must leave by the edge in $M_2$. But that is $ac$, and $P$ cannot follow it. So $P$ stops. Similarly, $P$ can only arrive at $x$ or $b$ by an edge in $M_2$, and must leave by $M_1$, i.e., by $xb$, which is not in $G$. Again, it must stop.
Finally, note that once $P$ visits a vertex, it can never visit it again. Suppose otherwise. Let $v$ be the first vertex that $P$ visits twice. $v \ne c$, since the only way $P$ can reach $c$ a second time is by its $M_1$ edge. Let $u$ be the vertex on the other side of that edge. In order to take $uc$ to $c$, $P$ has to visit $u$ a second time, so $c$ cannot be the first. But then $u$ cannot be the first repeat visit either, since $P$ cannot reach $u$ again from its $M_1$ edge $uc$, since it cannot have visited $c$, and it cannot reach $u$ again from its $M_2$ edge, since the other vertex for that edge would have to be visited twice before $u$. This same logic recursively shows that no vertex can on $P$ can be the first visited twice. Therefore $P$ can never visit a vertex twice, and since it started out at $c$, $c$ cannot be its ending place.
Thus the only possible places that $P$ can end are $a, b, x$. (And note that contrary to your post, the proof does allow $a$ as a possible ending place.) And since $P$ cannot cycle, and cannot stop anywhere else, it has to encounter one of them.