Unfortunately, your penalty term is not correct.
When working with QUBO, penalties should be equal to zero for all feasible solutions to the problem.
The proper way express $x_i + x_j \leq 1$ as a penalty is writing it as
$\gamma x_i x_j$ where $\gamma$ is a positive penalty scaler (assuming you minimize). Note that if $x_i = 1$ and $x_j = 0$ (or vice versa) then $\gamma x_i x_j = 0$. So the penalty does not contribute to the objective function in any way, as it suppose to do.
However, if $x_i, x_j = 1$. Then, clearly your constraint is violated and you get a penalty of $\gamma x_i x_j = \gamma$.
Regarding the D-wave documentation about inequality constraints. Yes, this is indeed correct. In general, you need to have slack variables to keep your penalties to be zero for all feasible solutions. I can give a general example. Suppose we have inequality constraint
$$\tag{1} q^Tx \leq c$$
where $q \in \mathbb{Z}_+^n$ is a vector of coefficients and $x \in \{0,1\}^n$ is a vector of decision variables. To incorporate this into the objective function you convert (1) into a squared penalty
$$\gamma (q^Tx - z)^2$$
where $z$ is an integer slack variable such that $0 \leq z \leq c$. When the constraint (1) is satisfied $\gamma (q^Tx - z)^2 = 0$ because the slack variable takes on the value of $q^Tx$, i.e. $q^Tx = z$. However, if the constraint is violated, then we have $q^Tx > c$. Since $z \leq c$ we get a non-zero penalty term, $\gamma (q^Tx - z)^2>0$.
Now, our problem should be QUBO, but we have an integer, non-binary variable $z$! This means we need to express $z$ as a binary expansion such that $z \leq c$. This is a bit tricky but doable. I just give one possible way of doing it and it is not the best way of doing it! We can express $z$ as
$$z = \sum_{k=1}^c ky_k$$
where $y_k \in \{0,1\}$. This makes sure that $z$ can take on any integer value between $0$ and $c$. There are many other more efficient and clever ways of expressing $z$ such that $z \leq c$.
Note that this only applies to $q$ being an integer vector with positive entries. If it is rational, the situation becomes slightly more complicated. But the idea is the same.