Can every $n\times n$ symmetric matrix over $\mathbb{F}_2$ be decomposed into $$\sum_{v=1}^k v_i v_i^T$$ for vectors $v_1, \dots,v_k\in\mathbb{F}_2^n$ and integer $k$?
As far as I know, for symmetric real matrices this is true and these vectors are orthogonal, but I am not sure if these properties hold over finite fields.
This is definitely possible over $\Bbb{F}_2$. If $v$ is the column vector with $1$s at positions $i$ and $j$ and zeros elsewhere, $i<j$, then $vv^T$ has $1$s at positions $(i,i), (i,j),(j,i)$ and $(j,j)$. Similarly if $w$ has a single $1$ at position $i$, then $ww^T$ has a single $1$ on the diagonal.
We can write every symmetric matrix as a linear combination of these. Use vectors of the first type to get the non-diagonal entries, and then "fix" the diagonal with vectors of the second type.
On the other hand, asking for orthogonality is a bit strange given that over $\Bbb{F}_2$ vectors are often orthogonal to themselves. Anyway, to get the symmetric matrix $$ A=\pmatrix{0&1\cr1&0\cr} $$ you need to use all the three non-zero vectors $v_1,v_2,v_3\in\Bbb{F}_2^2$, and $k=3$ is the smallest possible value. The conclusion is that we cannot ask for the vectors $v_i$ to be linearly independent.
At this time I don't want to say anything about the minimal required $k$ that works for larger matrices :-)