The vector subspace $W$ will be spanned by states that can be labeled as follows:
\begin{align}
|\vec k\rangle = |k_1, k_2, \dots, k_n\rangle
\end{align}
where $k_i = 0$ tells you that there is no fermion present in "mode" $i$ and $k_i = 1$ tells you that there is one fermion present in "mode" $i$. There are $2^n$ such states because there are $2^n$ possible binary sequences of length $n$. These states are defined by acting an appropriate composite of creation operators on the ground state
$$
|k_1, k_2, \dots, k_n\rangle = (a_1^\dagger)^{k_1}(a_2^\dagger)^{k_2}\cdots (a_n^\dagger)^{k_n}|0, 0, \dots, 0\rangle
$$
where the ground state is defined as the state that is annihilated by every one of the annihilation operators $a_i$. It can be shown that the states $|\vec k\rangle$ are orthogonal; $\langle \vec k| \vec \ell\rangle = \delta_{\vec k, \vec \ell}$. If $W$ is a subspace of Hilbert space $V$ having dimension larger than $W$, then $W_\perp$ will be the set of all states $|\psi\rangle\in V$ that are orthogonal to every state in $W$; $\langle \psi|\vec k\rangle = 0$ for all sequences $\vec k$. This might happen if, for example, one considers a scenario in which the system actually has more than $n$ modes. Say there are actually $N$ possible modes a given fermion can occupy, so in reality the states in $V$ would be written as
$$
|k_1, k_2, \dots, k_n, k_{n+1}, \dots, k_N\rangle.
$$
Suppose, further, than one defines
$$
|k_1, k_2, \dots, k_n, 0, \dots, 0\rangle = |k_1, k_2, \dots, k_n\rangle
$$
so that $W$ is a subspace spanned only by the states in which there are no fermions in the higher modes $n+1, n+2, \dots N$. Then any state where there is a fermion in at least one of the higher modes will be in $W_\perp$. As an example, the following state will be in $W_\perp$:
$$
|\underbrace{0, 0, \dots, 0}_\text{first $n$ slots}, 1, 0, \dots, 0\rangle.
$$
