Say I have a matrix $A$ and consider $B$ of the form $A+vv^T+ww^T$ for some vectors $v,w$.
How do I have to chose $v$ and $w$ in order for $\text{rank}B=\text{rank}A-2$?
I was thinking that $v$ and $w$ should have something to do with eigenvectors of $A$ but I don't want to restrict $B$, I want any rank rank$A-2$ matrix $B$ to be constructible from this method.
I infer you are working over $\mathbb R$.
your stated goal of
is doomed to fail. Suppose I select $A$ to be rank 2 and nilpotent. Then the only possibility is $B=\mathbf 0$.
By subadditivity of rank, this means that $\mathbf v \neq \mathbf 0$ and $\mathbf w \neq \mathbf 0$.
Thus
$\mathbf 0 = B = A+\mathbf {vv}^T+\mathbf {ww}^T$ but
$0 = \text{trace}\big(\mathbf 0\big) = \text{trace}\big( B\big) = \text{trace}\big(A+\mathbf {vv}^T+\mathbf {ww}^T\big) = 0 + \text{trace}\big(\mathbf {vv}^T\big) + \text{trace}\big(\mathbf {ww}^T\big)\gt 0 $
more generally: if you instead try for
$\mathbf 0 = B = A+ \alpha_1\mathbf {vv}^T+\alpha_2\mathbf {ww}^T$
you'll find this is still impossible, e.g. because it implies a symmetric matrix given by $-\big(\alpha_1\mathbf {vv}^T+\alpha_2\mathbf {ww}^T\big)$ is equal to $A$ which is (orthogonally similar to) a non-zero strictly upper triangular matrix.