If $\mathcal M$ is $\omega$-small, then many $\mathcal J^{\mathcal M}_\beta$ may think that (a large fragment of $\mathsf{ZF}$ holds and) there are plenty of Woodin cardinals. What matters is that for any such $\beta$ there is a larger $\tau$ where we see that this is no longer the case.

This may happen for a variety of reasons: Maybe the relevant cardinals got collapsed along the way. Or maybe they remain cardinals but new subsets $A$ got added, for which we do not have witnesses to $A$-strongness. Or maybe the previous witnesses for varying sets $A$ do not actually give rise to $A$-strong embeddings of $\mathcal J^{\mathcal M}_\tau$.

The only serious requirement is that $\tau$ is reached before we arrive at an active stage (one where we add a extender to the sequence).

Properly small mice have the additional requirement that we in fact certify that no cardinals are Woodin by the time we reach the top of the mouse.

Perhaps a simple-minded analogy may help: (Assume $V=L$ if you want.) If $\kappa$ is inaccessible, and $X\prec L_\kappa$ is countable, then the collapse of $X$ is an $L_\beta$ with $\beta$ countable such that $L_\beta$ is a model of $\mathsf{ZFC}$. However, there is an $\alpha$ with $\beta<\alpha<\omega_1$ such that $L_\alpha$ sees that $\beta$ is countable and, of course, by the time we reach $\omega_1$ we have certified that no smaller ordinal (other than $\omega$) is a cardinal.

If $\mathcal M$ is $\omega$-small, then many $\mathcal J^{\mathcal M}_\beta$ may think that (a large fragment of $\mathsf{ZF}$ holds and) there are plenty of Woodin cardinals. What matters is that for any such $\beta$ there is a larger $\tau$ where we see that this is no longer the case.

This may happen for a variety of reasons: Maybe the relevant cardinals got collapsed along the way. Or maybe they remain cardinals but new subsets $A$ got added, for which we do not have witnesses to $A$-strongness. Or maybe the previous witnesses for various sets $A$ do not actually give rise to $A$-strong embeddings of $\mathcal J^{\mathcal M}_\tau$.

The only serious requirement is that $\tau$ must be reached before we arrive at an active stage (one where we add an extender to the sequence).

Properly small mice have the additional requirement that we in fact certify that no cardinals are Woodin by the time we reach the height of the mouse.

Perhaps a simple-minded analogy may help: (Assume $V=L$ if you want.) If $\kappa$ is inaccessible, and $X\prec L_\kappa$ is countable, then the collapse of $X$ is an $L_\beta$ with $\beta$ countable such that $L_\beta$ is a model of $\mathsf{ZFC}$. However, there is an $\alpha$ with $\beta<\alpha<\omega_1$ such that $L_\alpha$ sees that $\beta$ is countable and, of course, by the time we reach $\omega_1$ we have certified that no smaller ordinal (past $\omega$) is a cardinal.

If $\mathcal M$ is $\omega$-small, then many $\mathcal J^{\mathcal M}_\beta$ may think that (a large fragment of $\mathsf{ZF}$ holds and) there are plenty of Woodin cardinals. What matters is that for any such $\beta$ there is a larger $\tau$ where we see that this is no longer the case.

This may happen for a variety of reasons: Maybe the relevant cardinals got collapsed along the way. Or maybe they remain cardinals but new subsets $A$ got added, for which we do not have witnesses to $A$-strongness. Or maybe the previous witnesses for varying sets $A$ do not actually give rise to $A$-strong embeddings of $\mathcal J^{\mathcal M}_\tau$.

The only serious requirement is that $\tau$ is reached before we arrive at an active stage (one where we add a extender to the sequence).

Properly small mice have the additional requirement that we in fact certify that no cardinals are Woodin by the time we reach the top of the mouse.

Perhaps a simple-minded analogy may help: (Assume $V=L$ if you want.) If $\kappa$ is inaccessible, and $X\prec L_\kappa$ is countable, then the collapse of $X$ is an $L_\beta$ with $\beta$ countable such that $L_\beta$ is a model of $\mathsf{ZFC}$. However, there is an $\alpha$ with $\beta<\alpha<\omega_1$ such that $L_\alpha$ sees that $\beta$ is countable and, of course, by the time we reach $\omega_1$ we have certified that no smaller ordinal is a cardinal.

If $\mathcal M$ is $\omega$-small, then many $\mathcal J^{\mathcal M}_\beta$ may think that (a large fragment of $\mathsf{ZF}$ holds and) there are plenty of Woodin cardinals. What matters is that for any such $\beta$ there is a larger $\tau$ where we see that this is no longer the case.

This may happen for a variety of reasons: Maybe the relevant cardinals got collapsed along the way. Or maybe they remain cardinals but new subsets $A$ got added, for which we do not have witnesses to $A$-strongness. Or maybe the previous witnesses for varying sets $A$ do not actually give rise to $A$-strong embeddings of $\mathcal J^{\mathcal M}_\tau$.

The only serious requirement is that $\tau$ is reached before we arrive at an active stage (one where we add a extender to the sequence).

Properly small mice have the additional requirement that we in fact certify that no cardinals are Woodin by the time we reach the top of the mouse.

Perhaps a simple-minded analogy may help: (Assume $V=L$ if you want.) If $\kappa$ is inaccessible, and $X\prec L_\kappa$ is countable, then the collapse of $X$ is an $L_\beta$ with $\beta$ countable such that $L_\beta$ is a model of $\mathsf{ZFC}$. However, there is an $\alpha$ with $\beta<\alpha<\omega_1$ such that $L_\alpha$ sees that $\beta$ is countable and, of course, by the time we reach $\omega_1$ we have certified that no smaller ordinal (other than $\omega$) is a cardinal.