It's quite possible for the window property to fail for a ``balanced'' sequence (where 1 more 1's occur than 0's in a period of length $2^n-1$). For example
a sequence of length $2^{n}-1$ made up of $2^{n-1}$ consecutive 1's followed by $2^{n-1}-1$ zeroes, taken as periodically repeating will have $2^{n-1}-n+1$ all 1 windows of length $n$ and $2^{n-1}-n+2$ all 0 windows of length $n$ with the remainder being windows of the form $$w_k=\underbrace{1111\cdots1}_{k~times}\underbrace{0000\cdots0}_{n-k~times}$$ for $k=1,2,\ldots,n-2$ and the complement of $w_k$ for $k=1,2,\ldots,n-1.$

The opposite, however is not possible. To see this take a sequence with the window property, and imagine fixing any bit position in the window as the window is slid along a full period of the sequence (with no loss of generality we can take this position to be the leftmost bit in the window, since we can go around the full cycle with any starting position we want in the cyclic sequence) . Now the windows range over the nonzero $n-$tuples over $\{0,1\}$ in a manner so that the consecutive windows are compatible sequences i.e.,
$$
x_k x_{k+1} \cdots x_{k+n-1}~~~\textrm{is followed by }~~~~x_k x_{k+1} \cdots x_{k+n-1}.
$$
Since there is a one to one correspondence between these sequence of windows and the nonzero $n-$tuples by the window property, there is a one to one correspondence inherited by any fixed bit position, say the first, and a collection of $2^{n-1}-1$ 0's and $2^{n-1}$ 1's must be traced by this point by straightforward counting.

Put it another way, any sequence of period $2^n-1$ with the $n-$window property also has the $n-a$window property for any $a=1,2,\ldots, n-1$ and in one period each nonzero $(n-a)-$tuple occurs $2^{a}$ times while the zero $(n-a)-$tuple occurs $2^{a}-1$ times.