$V, W$ are finite dimensional $\mathbb{K}$-Vectorspaces. $\,\,f: V \rightarrow W$ is an Isomorphism. $\,\,q_v \in \text{Quad}(V), \,\,q_w \in \text{Quad}(W)$.
It is known, that:
$$ q_v(v_1 + v_2) - q_v(v_1) - q_v(v_2) = (q_w \circ f)(v_1+v_2) - (q_w \circ f)(v_1) - (q_w \circ f)(v_2) $$ for every $v_1, v_2 \in V$. How can one prove that $q_v = q_w \circ f$? My first guess was setting $v_1$ or $v_2$ equal to zero, but then the equation collapses into $0 = 0$.
Try setting $v_2 = v_1 = v$. Then your equality says:
$$ \begin {align} q_V(2v) - 2q_V(v) &= q_W(f(2v)) - 2q_W(f(v)) \\ 2q_V(v) &= 2q_W(f(v)) \end {align} $$