It's not quite clear to me, but I think you mean there is some $k \in [0, \ldots, N+1-\alpha]$ such that
$\sum_{j=0}^M x_{ij} = 1$ for $i = k, k+1, \ldots, k+\alpha - 1$
while $\sum_{j=1}^M x_{ij} = 0$ for all other $i$. You can model this with
binary variables $t_i$, $i=0 \ldots N+1-\alpha$, and constraints
$$
\eqalign{\sum_{i=0}^{N+1-\alpha} t_i & = 1 \cr
\sum_{j=0}^M x_{ij} - \sum_{k=\max(0,i+1-\alpha)}^{\min(i,N+1-\alpha)} t_k &= 0
\text{ for each $ i = 0 \ldots N$}\cr}$$
The following assumes that $y_{ik} \geq 0$.
First, let's introduce a new set of variables $z_{ijk\ell}$:
\begin{gathered}
\sum_{i,j,k} z_{ijkl} \leq I_\ell \qquad \forall \ell \\
y_{ik} \cdot x_{ijk\ell} = z_{ijk\ell} \qquad \forall i,j,k,\ell
\end{gathered}
I hope you don't mind that for brevity I've dropped the index sets. This is equivalent (in $x$ and $y$) to
\begin{gathered}
\sum_{i,j,k} z_{ijkl} \leq I_\ell \qquad \forall \ell \\
y_{ik} \cdot x_{ijk\ell} \leq z_{ijk\ell} \qquad \forall i,j,k,\ell
\end{gathered}
This works because when the original inequality is feasible, we can always choose to have equality hold: but if there is slack in that inequality, we could distribute that slack across the values of $z_{ijk\ell}$ without changing equivalence.
Now that we've done that, we can use this equivalence:
$$y_{ik} \geq 0, ~~ y_{ik} \cdot x_{ijk\ell} \leq z_{ijk\ell} \quad\Longleftrightarrow\quad
0 \leq z_{ijk\ell}, ~~ y_{ik} - v_i ( 1 - x_{ijk\ell} ) \leq z_{ijk\ell}$$
If $x_{ijk\ell}=0$, then $z_{ijk\ell}\geq 0$ will be active. If $x_{ijk\ell}=1$, then the second constraint will be active, and it will be equivalent to $y_{ik} \leq z_{ijk\ell}$.
So the final conversion is
\begin{gathered}
\sum_{i,j,k} z_{ijkl} \leq I_\ell \qquad \forall \ell \\
y_{ik} - v_i (1- x_{ijk\ell}) \leq z_{ijk\ell} \qquad \forall i,j,k,\ell \\
0 \leq z_{ijk\ell} \qquad \forall i,j,k,\ell
\end{gathered}
Best Answer
Each $\alpha x_i$ term can be reformulated using a continuous variable $\beta_i$ and three constraints as follows:
$\beta_i \leq \alpha \\ \beta_i \leq x_i \\ \beta_i \geq \alpha + x_i - 1$
Reformulate the $\alpha y_j$ terms the same way using another continuous variable, say $\gamma_j$.
Then, use the big-M method from integer programming to handle the "either-or" constraint that you want to enforce, using a binary variable $\delta$, a sufficiently large constant $M$, and two constraints as follows:
$\sum_j \gamma_j - M\delta \leq \sum_i \beta_i \leq \sum_j \gamma_j + M\delta \\ \sum_i \beta_i \leq M(1-\delta)$