It is not necessarily that each path $P_i$ contains exactly one of $\{v_1, \cdots, v_k\}$. A counterexample is as follows.
Counterexample. The following figure shows a $2$-edge-connected graph, where $v$ is green, $v_1$ is blue and $v_2$ is red. As you can see, every path from $v$ to $v_2$ must pass $v_2$.
However, it is true that $P_i$ and $P_j$ are edge-disjoint if $i \neq j$. To prove this, you can construct a new graph $G'$ by adding a new vertex $w$ and $k$ new edges, namely $(w, v_1)$, $(w, v_2)$, $\cdots$ and $(w, v_k)$, to $G$. It is easy to know that $G'$ is $k$-edge connected. By Menger's theorem, there exists $k$ edge-disjoint paths from $v$ to $w$.
If we remove $w$ from each of these paths, the paths are from $v$ to each $v_i$ in $G$ and are edge-disjoint.
For each vertex of the graph, the level of the vertex is the length of the shortest directed path from $s$ to that vertex. So $s$ has level $0$, $v_{2l-1}$ & $v_{2l}$ have level $l$, and $t$ has level $k+1$. Note that each directed path from $s$ to $t$ much contain exactly one vertex from each level.
Let $s=u_0 \to u_1 \to u_2 \to\ldots \to u_k \to u_{k+1}=t$ be a directed path. Note that $u_i$ has level $i$. For each $u_i$ ($0\le i \le k-1$), there are always two possible vertices of level $i+1$ and $(u_i,u_{i+1})$ is always a directed edge in $G$. Also there is only one choice after $u_k$. Therefore, the number of such paths is $2^k$.
I also thought about bof's proposition to consider un-directed paths. Here is a proof.
Note that for an un-directed path $s=u_0\to u_1\to u_2\to\ldots \to u_r \to u_{r+1}=t$, there is at most one possible backward movement at each $u_i$ and if $u_{i+1}$ is obtained by a backward movement, $u_{i+1}$ must go to $u_{i+2}$ with a unique forward movement. Then, we cannot make another backward movement, so we have to perform another forward movement. Let's call such a backward-followed-by-forward-twice sequence a looping. Observe that two loopings cannot be done consecutively.
Observe also that if we are not at a vertex where a backward movement has just been performed, there are always two possible choices to move forward (unless you are at $v_{2k-1}$ or $v_{2k}$, where the only forward movement is to go to $t$). Therefore, our path can be represented by a binary sequence of length $k+1$, where $0$ is a forward movement and $1$ is a looping, such that the sequence starts with two $0$, and no two $1$ occur successively.
There are $F_{k+1}$ such binary sequences (since removing the two $0$ at the beginning, you are in the situation of this question). Here $F_k$ is the $k^\text{th}$ Fibonacci number with $F_0=0$ and $F_1=1$. There are $k$ forward movements each having two choices (recalling that the final forward movement has only one possible choice). This gives you a factor $2^k$. Therefore, there are $2^kF_{k+1}$ un-directed paths from $s$ to $t$.
Best Answer
https://en.wikipedia.org/wiki/Path_(graph_theory)
A path is a sequence of edges which connect a sequence of vertices which, by most definitions, are all distinct from one another.
So $v_1$ would not appear twice in a path (we would call that a cycle), but you'd have to name the edges to define a path.
$v_1v_2$ would equal $v_2v_1$ since it is the same edge.
$v_1v_2v_3$ would be a different path then $v_1v_3v_2$ since the edge sequence is different.
(By these definitions I'd say you have 4 distinct paths)