This is a comment to Daniel Loughran's answer; I would like to add some more details on obstruction to lifting $PO(n)$-cocycles to $O(n)$-cocycles (at least in the case where $k$ has characteristic $\neq 2$).

As discussed in the Wikipedia article, there is an extension of algebraic groups
$$ 1\to \mathbb{Z}/2=\{\pm I\}\to O(Q)\to PO(Q)\to 1, $$ which is split in the odd case and non-split in the even case. Then there is an exact sequence in group cohomology $$H^1(k,O(Q))\to H^1(k,PO(Q))\to H^2(k,\mathbb{Z}/2), $$ see e.g. the Galois cohomology book of Serre. Since the sequence is split in case $Q$ is odd-dimensional, the element $\sigma\in H^1(k,PO(Q))$ classifying the form maps trivially to $H^2(k,\mathbb{Z}/2)$ and so the map $H^1(k,O(Q))\to H^1(k,PO(Q))$ is surjective. In the case where $Q$ is even-dimensional (which corresponds to the case $\mathbb{P}^{2n+1}$ mentioned in Daniel Loughran's answer), the extension is non-split but it is split locally in the étale topology. Therefore, the extension class lives in $H^2_{\operatorname{et}}(k,\mathbb{Z}/2)$.

This cohomology group has several interpretations. By the Merkurjev-Suslin theorem (a special case of the Milnor conjecture), there is an isomorphism $H^2_{\operatorname{et}}(k,\mathbb{Z}/2)\cong K^M_2(k)/2K^M_2(k)$. By a theorem of Merkurjev, $K^M_2(k)/2K^M_2(k)\cong {}_2Br(k)$, explaining the appearance of $2$-torsion in the Brauer group in Daniel Loughran's answer. (A possible reference for these would be the book on central simple algebras by P. Gille and T. Szamuely; alternatively, check out survey papers on the Milnor conjecture.) So for each form of the quadric $Q$ (parametrized by an element $\sigma\in H^1(k,PO(Q))$) there is an associated obstruction class in $H^2(k,\mathbb{Z}/2)\cong K^M_2(k)/2\cong {}_2Br(k)$ whose triviality is equivalent to the form being a quadric.

Probably (but I have not checked this) Daniel Loughran's example can be explained as follows: each of the conics corresponds to a class in $H^1(k,\mathbb{Z}/2)$, and the obstruction class is the cup-product of these two classes.

This is a comment to Daniel Loughran's answer; I would like to add some more details on obstruction to lifting $PO(n)$-cocycles to $O(n)$-cocycles (at least in the case where $k$ has characteristic $\neq 2$).

As discussed in the Wikipedia article, there is an extension of algebraic groups
$$ 1\to \mathbb{Z}/2=\{\pm I\}\to O(Q)\to PO(Q)\to 1, $$ which is split in the odd case and non-split in the even case. Then there is an exact sequence in group cohomology $$H^1(k,O(Q))\to H^1(k,PO(Q))\to H^2(k,\mathbb{Z}/2), $$ see e.g. the Galois cohomology book of Serre. Since the sequence is split in case $Q$ is odd-dimensional, the element $\sigma\in H^1(k,PO(Q))$ classifying the form maps trivially to $H^2(k,\mathbb{Z}/2)$ and so the map $H^1(k,O(Q))\to H^1(k,PO(Q))$ is surjective. In the case where $Q$ is even-dimensional (which corresponds to the case $\mathbb{P}^{2n+1}$ mentioned in Daniel Loughran's answer), the extension is non-split but it is split locally in the étale topology. Therefore, the extension class lives in $H^2_{\operatorname{et}}(k,\mathbb{Z}/2)$.

This cohomology group has several interpretations. By the Merkurjev-Suslin theorem (a special case of the Milnor conjecture), there is an isomorphism $H^2_{\operatorname{et}}(k,\mathbb{Z}/2)\cong K^M_2(k)/2K^M_2(k)$. By a theorem of Merkurjev, $K^M_2(k)/2K^M_2(k)\cong {}_2Br(k)$, explaining the appearance of $2$-torsion in the Brauer group in Daniel Loughran's answer. (A possible reference for these would be the book on central simple algebras by P. Gille and T. Szamuely; alternatively, check out survey papers on the Milnor conjecture.) So for each form of the quadric $Q$ (parametrized by an element $\sigma\in H^1(k,PO(Q))$) there is an associated obstruction class in $H^2(k,\mathbb{Z}/2)\cong K^M_2(k)/2\cong {}_2Br(k)$ whose triviality is equivalent to the form being a quadric.

This is a comment to Daniel Loughran's answer; I would like to add some more details on obstruction to lifting $PO(n)$-cocycles to $O(n)$-cocycles (at least in the case where $k$ has characteristic $\neq 2$).

As discussed in the Wikipedia article, there is an extension of algebraic groups
$$ 1\to \mathbb{Z}/2=\{\pm I\}\to O(Q)\to PO(Q)\to 1, $$ which is split in the odd case and non-split in the even case. Then there is an exact sequence in group cohomology $$H^1(k,O(Q))\to H^1(k,PO(Q))\to H^2(k,\mathbb{Z}/2), $$ see e.g. the Galois cohomology book of Serre. Since the sequence is split in case $Q$ is odd-dimensional, the element $\sigma\in H^1(k,PO(Q))$ classifying the form maps trivially to $H^2(k,\mathbb{Z}/2)$ and so the map $H^1(k,O(Q))\to H^1(k,PO(Q))$ is surjective. In the case where $Q$ is even-dimensional (which corresponds to the case $\mathbb{P}^{2n+1}$ mentioned in Daniel Loughran's answer), the extension is non-split but it is split locally in the étale topology. Therefore, the extension class lives in $H^2_{\operatorname{et}}(k,\mathbb{Z}/2)$.

This cohomology group has several interpretations. By the Merkurjev-Suslin theorem (a special case of the Milnor conjecture), there is an isomorphism $H^2_{\operatorname{et}}(k,\mathbb{Z}/2)\cong K^M_2(k)/2K^M_2(k)$. By a theorem of Merkurjev, $K^M_2(k)/2K^M_2(k)\cong {}_2Br(k)$, explaining the appearance of $2$-torsion in the Brauer group in Daniel Loughran's answer. So for each form of the quadric $Q$ (parametrized by an element $\sigma\in H^1(k,PO(Q))$) there is an associated obstruction class in $H^2(k,\mathbb{Z}/2)\cong K^M_2(k)/2\cong {}_2Br(k)$ whose triviality is equivalent to the form being a quadric.

Probably (but I have not checked this) Daniel Loughran's example can be explained as follows: each of the conics corresponds to a class in $H^1(k,\mathbb{Z}/2)$, and the obstruction class is the cup-product of these two classes.

This is a comment to Daniel Loughran's answer; I would like to add some more details on obstruction to lifting $PO(n)$-cocycles to $O(n)$-cocycles (at least in the case where $k$ has characteristic $\neq 2$).

As discussed in the Wikipedia article, there is an extension of algebraic groups
$$ 1\to \mathbb{Z}/2=\{\pm I\}\to O(Q)\to PO(Q)\to 1, $$ which is split in the odd case and non-split in the even case. Then there is an exact sequence in group cohomology $$H^1(k,O(Q))\to H^1(k,PO(Q))\to H^2(k,\mathbb{Z}/2), $$ see e.g. the Galois cohomology book of Serre. Since the sequence is split in case $Q$ is odd-dimensional, the element $\sigma\in H^1(k,PO(Q))$ classifying the form maps trivially to $H^2(k,\mathbb{Z}/2)$ and so the map $H^1(k,O(Q))\to H^1(k,PO(Q))$ is surjective. In the case where $Q$ is even-dimensional (which corresponds to the case $\mathbb{P}^{2n+1}$ mentioned in Daniel Loughran's answer), the extension is non-split but it is split locally in the étale topology. Therefore, the extension class lives in $H^2_{\operatorname{et}}(k,\mathbb{Z}/2)$.

This cohomology group has several interpretations. By the Merkurjev-Suslin theorem (a special case of the Milnor conjecture), there is an isomorphism $H^2_{\operatorname{et}}(k,\mathbb{Z}/2)\cong K^M_2(k)/2K^M_2(k)$. By a theorem of Merkurjev, $K^M_2(k)/2K^M_2(k)\cong {}_2Br(k)$, explaining the appearance of $2$-torsion in the Brauer group in Daniel Loughran's answer. (A possible reference for these would be the book on central simple algebras by P. Gille and T. Szamuely; alternatively, check out survey papers on the Milnor conjecture.) So for each form of the quadric $Q$ (parametrized by an element $\sigma\in H^1(k,PO(Q))$) there is an associated obstruction class in $H^2(k,\mathbb{Z}/2)\cong K^M_2(k)/2\cong {}_2Br(k)$ whose triviality is equivalent to the form being a quadric.

Probably (but I have not checked this) Daniel Loughran's example can be explained as follows: each of the conics corresponds to a class in $H^1(k,\mathbb{Z}/2)$, and the obstruction class is the cup-product of these two classes.