Here's a proof assuming $n\ge 3,n\neq 8$ that the Lie algebras are isomorphic only when the quadratic forms are equivalent up to rescaling (I assume $K$ has characteristic zero and fix an algebraically closed extension $C$).

Let $f:\mathfrak{so}(\phi)\to \mathfrak{so}(\psi)$ be a $K$-defined isomorphism. We can assume that both $\mathfrak{so}(\phi)$ and $\mathfrak{so}(\psi)$ are $K$-defined subalgebras of $\mathfrak{sl}_n$ preserving a nondegenerate quadratic form on the $n$-dimensional space. Then both are $C$-conjugate to $\mathfrak{so}(n)$. Since I assume $n\neq 8$ odd, over $C$, all automorphisms of $\mathfrak{so}(n)$ can be realized by some element of $\mathrm{GL}_n(C)$ (actually, of $\mathrm{O}_n(C)$). It follows that $f$ can be realized by a conjugation, namely there exists $A\in\mathrm{GL}_n(C)$ satisfying: $f(g)A=Ag$ for all $g\in \mathfrak{so}(\phi)$. The set of $A$ satisfying this condition is a $K$-defined linear subspace on which the determinant map does not vanish; hence it contains a $K$-point with nonzero determinant. That is, $A$ can be found in $GL_n(K)$. Hence $\mathfrak{so}(\psi)$ preserves the $K$-defined quadratic form $x\mapsto \phi'(x):=\phi(A^{-1}x)$. Since the set of $K$-defined invariant forms is 1-dimensional (because the standard representation of $\mathfrak{so}(\psi)$ is absolutely irreducible), it follows that $\phi'$ and $\psi$ are collinear.

Note that the result in terms of Lie algebras is false for $n=2$ (the Lie algebra is the one-dimensional Lie algebra which is unique over $K$, but $\mathrm{SO}(\phi)$ is a 1-dimensional torus, which can be split or not. I don't know what's going on for $n=8$.

Here's a proof assuming $n\ge 3,n\neq 8$ that the Lie algebras are isomorphic only when the quadratic forms are equivalent up to rescaling (I assume $K$ has characteristic zero and fix an algebraically closed extension $C$).

Let $f:\mathfrak{so}(\phi)\to \mathfrak{so}(\psi)$ be a $K$-defined isomorphism. We can assume that both $\mathfrak{so}(\phi)$ and $\mathfrak{so}(\psi)$ are $K$-defined subalgebras of $\mathfrak{sl}_n$ preserving a nondegenerate quadratic form on the $n$-dimensional space. Then both are $C$-conjugate to $\mathfrak{so}(n)$. Since I assume $n\neq 8$ odd, over $C$, all automorphisms of $\mathfrak{so}(n)$ can be realized by some element of $\mathrm{GL}_n(C)$ (actually, of $\mathrm{O}_n(C)$). It follows that $f$ can be realized by a conjugation, namely there exists $A\in\mathrm{GL}_n(C)$ satisfying: $f(g)A=Ag$ for all $g\in \mathfrak{so}(\phi)$. The set of $A$ satisfying this condition is a $K$-defined linear subspace on which the determinant map does not vanish; hence it contains a $K$-point with nonzero determinant. That is, $A$ can be found in $GL_n(K)$. Hence $\mathfrak{so}(\psi)$ preserves the $K$-defined quadratic form $x\mapsto \phi'(x):=\phi(A^{-1}x)$. Since the set of $K$-defined invariant forms is 1-dimensional (because the standard representation of $\mathfrak{so}(\psi)$ is absolutely irreducible), it follows that $\phi'$ and $\psi$ are collinear.

I don't know what's going on for $n=8$.

For $n=2$, while there's only one 1-dimensional Lie algebra over $K$ so it's not enough to classify. Nevertheless it's still true that two quadratic forms are $K$-isomorphic up to rescaling [equivalently, have same determinant in $K^*/(K^*)^2$] iff they have isogenous SO(-). The point is that for 1-dimensional $K$-tori, $K$-isogenous is the same as $K$-isomorphic, and we can run the same proof as the above Lie-algebra-theoretic one, where we need to use the fact that every automorphism of $\mathrm{SO}_2(C)$ can be realized by some element of $\mathrm{GL}_2$. Actually this latter proof works for all $n\ge 2$ to show that if $\mathrm{SO}(\phi)$ and $\mathrm{SO}(\psi)$ are $K$-isomorphic then $\phi$ and $\psi$ are $K$-equivalent up to rescaling.

Here's a proof for odd dimension $n\ge 3$ that the Lie algebras are isomorphic only when the quadratic forms are equivalent up to rescaling (I assume $K$ has characteristic zero and fix an algebraically closed extension $C$).

Let $f:\mathfrak{so}(\phi)\to \mathfrak{so}(\psi)$ be a $K$-defined isomorphism. We can assume that both $\mathfrak{so}(\phi)$ and $\mathfrak{so}(\psi)$ are $K$-defined subalgebras of $\mathfrak{sl}_n$ preserving a nondegenerate quadratic form on the $n$-dimensional space. Then both are $C$-conjugate to $\mathfrak{so}(n)$. Since I assume $n$ odd, over $C$, all automorphisms of $\mathfrak{so}(n)$ are inner. It follows that $f$ can be realized by a conjugation, namely there exists $A\in\mathrm{GL}_n(C)$ satisfying: $f(g)A=Ag$ for all $g\in \mathfrak{so}(\phi)$. The set of $A$ satisfying this condition is a $K$-defined linear subspace on which the determinant map does not vanish; hence it contains a $K$-point with nonzero determinant. That is, $A$ can be found in $GL_n(K)$. Hence $\mathfrak{so}(\psi)$ preserves the $K$-defined quadratic form $x\mapsto \phi'(x):=\phi(A^{-1}x)$. Since the set of $K$-defined invariant forms is 1-dimensional (because the standard representation of $\mathfrak{so}(\psi)$ is absolutely irreducible), it follows that $\phi'$ and $\psi$ are collinear.

A similar argument still works when the outer automorphisms of $\mathfrak{so}(\phi)$ can be realized over $K$, that is, setting $G=\mathrm{Aut}(\mathfrak{so}(\phi))$, when the natural homomorphism $G_K\to (G/G^0)(C)$ is onto. (Recall that the latter group has order 1 for odd $n$, order 2 for $n\ge 4$ except $n=8$, and order 6 for $n=8$.) I don't know examples for which it's not the case but I haven't yet thought about it.

Here's a proof assuming $n\ge 3,n\neq 8$ that the Lie algebras are isomorphic only when the quadratic forms are equivalent up to rescaling (I assume $K$ has characteristic zero and fix an algebraically closed extension $C$).

Let $f:\mathfrak{so}(\phi)\to \mathfrak{so}(\psi)$ be a $K$-defined isomorphism. We can assume that both $\mathfrak{so}(\phi)$ and $\mathfrak{so}(\psi)$ are $K$-defined subalgebras of $\mathfrak{sl}_n$ preserving a nondegenerate quadratic form on the $n$-dimensional space. Then both are $C$-conjugate to $\mathfrak{so}(n)$. Since I assume $n\neq 8$ odd, over $C$, all automorphisms of $\mathfrak{so}(n)$ can be realized by some element of $\mathrm{GL}_n(C)$ (actually, of $\mathrm{O}_n(C)$). It follows that $f$ can be realized by a conjugation, namely there exists $A\in\mathrm{GL}_n(C)$ satisfying: $f(g)A=Ag$ for all $g\in \mathfrak{so}(\phi)$. The set of $A$ satisfying this condition is a $K$-defined linear subspace on which the determinant map does not vanish; hence it contains a $K$-point with nonzero determinant. That is, $A$ can be found in $GL_n(K)$. Hence $\mathfrak{so}(\psi)$ preserves the $K$-defined quadratic form $x\mapsto \phi'(x):=\phi(A^{-1}x)$. Since the set of $K$-defined invariant forms is 1-dimensional (because the standard representation of $\mathfrak{so}(\psi)$ is absolutely irreducible), it follows that $\phi'$ and $\psi$ are collinear.

Note that the result in terms of Lie algebras is false for $n=2$ (the Lie algebra is the one-dimensional Lie algebra which is unique over $K$, but $\mathrm{SO}(\phi)$ is a 1-dimensional torus, which can be split or not. I don't know what's going on for $n=8$.