I'm interested in matrices that are (i) square $n\times n$, (ii) has entries $0$ and $1$ only (iii) on each row, all the $1$'s are located a fixed distance $d<n$ from the main diagonal elements. What are these matrices called and can you please link to sources that discuss their properties? (Strictly speaking, I'm interested in $\frac{1}{2}$ times these matrices but I'll appreciate help either way.)
Examples: with $n=4,d=0$, we get the identity matrix $I_4$. With $n=4,d=1$, we have \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} and with $n=4$, $d=2$, we have \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}
It's helpful to consider only the case where $d \neq 0$, since $d = 0$ is a bit of a special case. Let $A(d,n)$ denote the matrix in which we have set $d$ equal to $d$ and $n$ equal to $n$ (so, your second example is $A(1,4)$).
Two interesting decompositions:
First, let $e_1,\dots,e_n$ denote the canonical basis. If we set $P$ equal to the permutation matrix $$ P = \pmatrix{e_1 & e_{1 + d} & e_{1 + 2d} & \cdots & e_2 & e_{2 + d} & e_{2 + 2d} \cdots & \cdots} $$ Then we find that $$ P^TA(d,n)P = \pmatrix{A(1,n_1)\\ & A(1,n_2) && \\ &&\ddots \\ &&& A(1,n_k)} $$ where $n_i$ is the length of the sequence $i,i + d, i + 2d, \dots, 1 + n_id$; $n_i$ is the largest integer such that $i + n_i d \leq n$ (that is, $n_i = \lfloor (n - i)/d\rfloor$).
We can decompose $A(1,n)$ in a more informative way by noting that if we set $$ Q = \begin{cases} \pmatrix{e_1 & e_3 & \cdots & e_{n-1} & e_2 & e_4 & \cdots & e_{n}} & \text{n is even}\\ \pmatrix{e_1 & e_3 & \cdots & e_{n} & e_2 & e_4 & \cdots & e_{n-1}} & \text{n is odd} \end{cases} $$ then we find that $$ Q^TA(1,n)Q = \pmatrix{0 & B\\B^T & 0} $$ where $B$ is the identity matrix of size $n/2$ when $n$ is even, and the identity matrix of size $(n-1)/2$ with an extra row of zeros when $n$ is odd.
With all that being said, your matrices are the adjacency matrices of the union of path graphs, which in particular means that they are all the adjacency matrices of bipartite graphs. We can conclude that the eigenvalues of $A$ will always come in $\pm$ pairs, so the positive signature of your bilinear form will be exactly equal to its negative signature.
It may also be useful to consider the full eigendecomposition of $A(1,n)$: since $A$ is tridiagonal and Toeplitz, we can quickly find that its eigenvalues are of the form $- \cos(k \pi/(n + 1))$ for $k = 1, \dots, n$. In particular, we see that $A(1,n)$ has no repeated eigenvalues and that it is singular (with nullity at most $1$) if and only if $n$ is odd.
Second, define $M$ by $$ M = \pmatrix{1 & 0\\ 0 & 1\\ \vdots & 0\\ 0 & \vdots & \cdots\\ 1 & 0\\ 0 & 1\\ \vdots & 0\\ & \vdots} $$ Where the $0 \cdots 0$ denotes $d-1$ consecutive zeros, and the final column ends in a $1$, with no trailing zeros. We can then express your matrix $A$ as $$ A = MM^T - D $$ where $D$ is a diagonal matrix of the form $D = \operatorname{diag}(1, \dots, 1,2 \dots , 2,1 ,\dots, 1)$. This decomposition was obtained using the incidence matrix of the associated graph.
This doesn't seem as immediately useful to me, but maybe you'll find something here.