How to prove LastToken problem is NP-complete
Here is a reduction from the Hamiltonian path. Given a graph $G=(V,E)$. Add a vertex $v_0$ to the graph and connect it to all vertices in the graph. Set $t(v_0)=2$. Set $t(u) = 1$ for all $u \neq v_0$.

Claim. The previous reduction is correct. Try to prove it formally as an exercise.

Edit. Here is a brief proof of correctness. We have to prove that the created instance is a yes-instance of your problem, if and only if the given instance is a yes-instance of the Hamiltonian path problem. First, let $G$ be a yes instance of the Hamiltonian path. Let $v_1, \dots v_n$ be a Hamiltonian cycle in the graph. Now we describe a winning strategy for the last-token, where in the step $i$ we take two tokens from the vertex $v_{i-1}$ and add one to the vertex $v_{i}$. It is easy to prove by induction, that in step $i+1$ all vertices $v_0, \dots v_{i-1}$ have no tokens in them, the vertex $v_i$ has two and all vertices $v_{i+1}, \dots v_n$ have exactly one token. In the $n-th$ step we take the two tokens left in $c_n$ and put one token in $v_0$ and we are done. Note that $v_i$ has always a common edge with $v_{i+1}$ since the are consecutive in a Hamiltonian path.

On the other hand, assuming the reduced instance is a yes-instance, we prove that the given graph is Hamiltonian. Since all but one vertex have exactly one token. We have only one choice for the first vertex. In each step we take two tokens from a vertex and add a token to another one. It is easy to prove inductively that in each step all but at most one vertex have at most one token in them. One vertex can have two tokens wich is the only choice for the next step. Using this claim, it is also easy to prove that to vertex can be chosen twice. Hence, if we were able to choose $n+1$ vertices, we must have a permutation of the $n$ vertices with $v_0$ starting in $v_0$ and this permutation must be a Hamiltonian path since we always have an edge between two consecutive choices in the game.