The $2^n \times 2^n$ Walsh-Hadamard matrix is given explicitly by
$$ M(u,v) = (-1)^{\langle u,v \rangle}, $$
where $u,v \in \{0,1\}^n$, and the inner product is given by
$$ \langle u,v \rangle = \sum_{i=1}^n u_i v_i. $$
To simplify matters, we will assume henceforth that all computations are modulo 2.
If we restrict to columns $v_1,\ldots,v_k$, then we see a row $\beta_1,\ldots,\beta_k$ whenever $$ \langle u, v_i \rangle = \beta_i, \quad i \in [k]. $$
We can think of this as a linear system in $u_1,\ldots,u_n \in \mathbb{F}_2$. If the system has rank $r$, then there are either no solutions, or $2^{n-r}$ solutions.
In your case, you are identifying rows of the forms $\beta_1,\ldots,\beta_k$ and $\beta_1+1,\ldots,\beta_k+1$. A row is of one of these forms whenever
$$
\langle u, v_i + v_k \rangle = \beta_i, \quad i \in [k-1].
$$
Again, this system either has no solutions, or $2^{n-r}$ solutions, where $r$ is the rank.
What this means is that every equivalence class of rows which appears at all, appears the same number of times.
In general, not all $2^{k-1}$ will appear. Indeed, there are only $2^n$ many rows, so if $k > n+1$ then this will certainly not be the case.
Another example is $n = 3$ and $S = \{0,\ldots,7\}$. Here only $4$ out of the $8$ possible equivalence classes appear, each of them appearing twice:
$$
\begin{bmatrix}
1&1&1&1\\
1&-1&1&-1\\
1&1&-1&-1\\
1&-1&-1&1\\
1&1&1&1\\
1&-1&1&-1\\
1&1&-1&-1\\
1&-1&-1&1
\end{bmatrix}
$$
