For simplicity, suppose $V$ is just a vector space (finite-dimensional, if you like). Let $\omega\in\Lambda^2 V^\vee$ (you start with a symmetric bilinear form, but combining it with the antisymmetric integral pairing of one-form gives an antisymmetric bilinear form). Then there are two forms associated to it: the constant two-form $\omega$ and the one-form $\eta$ whose value on a tangent vector $Y\in T_XV$ is $\omega(X,Y)$. Here $X\in V$ is regarded as a point of the manifold $V$, and $Y$ is regarded as a (constant) vector field on $V$.

Now let $X,Y\in V$ and regard them as (constant) vector fields on $V$. Then the usual formula for the differential gives \begin{align*} \mathrm d\eta(X,Y) &= X(\eta(Y)) - Y(\eta(X)) - \eta([X,Y])\\ &= \omega(X,Y) - \omega(Y,X) - 0\\ &= 2\omega(X,Y) \end{align*}

Alternatively, consider the ``Euler'' vector field $E$ whose value at a point $X\in V$ is $X\in T_XV\cong V$. Then by definition, $\eta = \iota_E\omega$. Since $\omega$ is closed, Cartan's formula gives $\mathrm d\eta = \mathcal L_E\omega = 2\omega$ since $E$ is the infinitesimal generator of scalar multiplication, so a constant coefficient $k$-form has weight $k$ under it.

More alternatively, suppose we consider some vector space $V$ over a field $K$ of characteristic $2$. Then there is still an exact sequence $(V\otimes V)^{C_2}\to V\otimes V\to (V\otimes V)_{C_2}$ of $GL_K(V)$-modules which essentially plays the role of the deRham complex of $V$, but it no longer splits $GL_K(V)$-equivariantly. This means that any construction which takes a constant coefficient antisymmetric form and produces a primitive for it must divide by $2$ or use an explicit basis (otherwise the formula would still make sense over $K$).

As to your calculation that $\mathrm d\eta = \dot\eta$, the only way I know how to make sense of the left-hand side is to say that $\eta = E$ is the Euler vector field and $\mathrm d$ is the canonical affine connection on the trivial tangent bundle of $V$. Thus $\mathrm d\eta$ should be a one-form with values in the tangent bundle (it is of course the canonical one-form). We do indeed have $\mathrm d\omega(\eta,\dot\eta) = \omega(\dot\eta,\dot\eta)$, but this is not the constant coefficient two-form determined by $\omega$: It takes two tangent vectors and first produces $\dot\eta\wedge\dot\eta(X,Y) = X\otimes Y - Y\otimes X$, then evaluates $\omega$ on it. Since $\omega$ was already antisymmetric, we get the same result twice, which explains the overall factor of $2$.

For simplicity, suppose $V$ is just a vector space (finite-dimensional, if you like). Let $\omega\in\Lambda^2 V^\vee$ (you start with a symmetric bilinear form, but combining it with the antisymmetric integral pairing of one-form gives an antisymmetric bilinear form). Then there are two forms associated to it: the constant two-form $\omega$ and the one-form $\xi_\omega$ whose value on a tangent vector $Y\in T_XV$ is $\omega(X,Y)$. Here $X\in V$ is regarded as a point of the manifold $V$, and $Y$ is regarded as a (constant) vector field on $V$.

Now let $X,Y\in V$ and regard them as (constant) vector fields on $V$. Then the usual formula for the differential gives \begin{align*} \mathrm d\xi_\omega(X,Y) &= X(\xi_\omega(Y)) - Y(\xi_\omega(X)) - \xi_\omega([X,Y])\\ &= \omega(X,Y) - \omega(Y,X) - 0\\ &= 2\omega(X,Y) \end{align*}

Alternatively, consider the ``Euler'' vector field $E$ whose value at a point $X\in V$ is $X\in T_XV\cong V$. Then by definition, $\xi_\omega = \iota_E\omega$. Since $\omega$ is closed, Cartan's formula gives $\mathrm d\xi_\omega = \mathcal L_E\omega = 2\omega$ since $E$ is the infinitesimal generator of scalar multiplication, so a constant coefficient $k$-form has weight $k$ under it.

More alternatively, suppose we consider some vector space $V$ over a field $K$ of characteristic $2$. Then there is still an exact sequence $(V\otimes V)^{C_2}\to V\otimes V\to (V\otimes V)_{C_2}$ of $GL_K(V)$-modules which essentially plays the role of the deRham complex of $V$, but it no longer splits $GL_K(V)$-equivariantly. This means that any construction which takes a constant coefficient antisymmetric form and produces a primitive for it must divide by $2$ or use an explicit basis (otherwise the formula would still make sense over $K$).

As to your calculation that $\mathrm d\eta = \dot\eta$, the only way I know how to make sense of the left-hand side is to say that $\eta = E$ is the Euler vector field and $\mathrm d$ is the canonical affine connection on the trivial tangent bundle of $V$. Thus $\mathrm d\eta$ should be a one-form with values in the tangent bundle (it is of course the canonical one-form). We do indeed have $\mathrm d\omega(\eta,\dot\eta) = \omega(\dot\eta,\dot\eta)$, but this is not the constant coefficient two-form determined by $\omega$: It takes two tangent vectors and first produces $\dot\eta\wedge\dot\eta(X,Y) = X\otimes Y - Y\otimes X$, then evaluates $\omega$ on it. Since $\omega$ was already antisymmetric, we get the same result twice, which explains the overall factor of $2$.

For simplicity, suppose $V$ is just a vector space (finite-dimensional, if you like). Let $\omega\in\Lambda^2 V^\vee$ (you start with a symmetric bilinear form, but combining it with the antisymmetric integral pairing of one-form gives an antisymmetric bilinear form). Then there are two forms associated to it: the constant two-form $\omega$ and the one-form whose value on a tangent vector $Y\in T_XV$ is $\omega(X,Y)$. Here $X\in V$ is regarded as a point of the manifold $V$, and $Y$ is regarded as a (constant) vector field on $V$.

Now let $X,Y\in V$ and regard them as (constant) vector fields on $V$. Then the usual rule for the differential gives \begin{align*} \mathrm d\eta(X,Y) &= X(\eta(Y)) - Y(\eta(X)) - \omega([X,Y])\\ &= \omega(X,Y) - \omega(Y,X) - 0\\ &= 2\omega(X,Y) \end{align*}

Alternatively, consider the ``Euler'' vector field $E$ whose value at a point $X\in V$ is $X\in T_XV\cong V$. Then by definition, $\eta = \iota_E\omega$. Since $\omega$ is closed, Cartan's formula gives $\mathrm d\eta = \mathcal L_E\omega = 2\omega$ since $E$ is the infinitesimal generator of scalar multiplication, so a constant coefficient $k$-form has weight $k$ under it.

More alternatively, suppose we consider some vector space $V$ over a field $K$ of characteristic $2$. Then there is still an exact sequence $(V\otimes V)^{C_2}\to V\otimes V\to (V\otimes V)_{C_2}$ of $GL_K(V)$-modules which essentially plays the role of the deRham complex of $V$, but it no longer splits $GL_K(V)$-equivariantly. This means that any construction which takes a constant coefficient antisymmetric form and produces a primitive for it must divide by $2$ or use an explicit basis (otherwise the formula would still make sense over $K$).

As to your calculation that $\mathrm d\eta = \dot\eta$, the only way to make sense of the left-hand side is to say that $\eta = E$ is the Euler vector field and $\mathrm d$ is the canonical affine connection on the trivial tangent bundle of $V$. Thus $\mathrm d\eta$ should be a one-form with values in the tangent bundle (it is of course the canonical one-form). We do indeed have $\mathrm d\omega(\eta,\dot\eta) = \omega(\dot\eta,\dot\eta)$, but this is not the constant coefficient two-form determined by $\omega$: It takes two tangent vectors and first produces $\dot\eta\wedge\dot\eta(X,Y) = X\otimes Y - Y\otimes X$, then evaluates $\omega$ on it. Since $\omega$ was already antisymmetric, we get the same result twice, which explains the overall factor of $2$.

For simplicity, suppose $V$ is just a vector space (finite-dimensional, if you like). Let $\omega\in\Lambda^2 V^\vee$ (you start with a symmetric bilinear form, but combining it with the antisymmetric integral pairing of one-form gives an antisymmetric bilinear form). Then there are two forms associated to it: the constant two-form $\omega$ and the one-form $\eta$ whose value on a tangent vector $Y\in T_XV$ is $\omega(X,Y)$. Here $X\in V$ is regarded as a point of the manifold $V$, and $Y$ is regarded as a (constant) vector field on $V$.

Now let $X,Y\in V$ and regard them as (constant) vector fields on $V$. Then the usual formula for the differential gives \begin{align*} \mathrm d\eta(X,Y) &= X(\eta(Y)) - Y(\eta(X)) - \eta([X,Y])\\ &= \omega(X,Y) - \omega(Y,X) - 0\\ &= 2\omega(X,Y) \end{align*}

Alternatively, consider the ``Euler'' vector field $E$ whose value at a point $X\in V$ is $X\in T_XV\cong V$. Then by definition, $\eta = \iota_E\omega$. Since $\omega$ is closed, Cartan's formula gives $\mathrm d\eta = \mathcal L_E\omega = 2\omega$ since $E$ is the infinitesimal generator of scalar multiplication, so a constant coefficient $k$-form has weight $k$ under it.

More alternatively, suppose we consider some vector space $V$ over a field $K$ of characteristic $2$. Then there is still an exact sequence $(V\otimes V)^{C_2}\to V\otimes V\to (V\otimes V)_{C_2}$ of $GL_K(V)$-modules which essentially plays the role of the deRham complex of $V$, but it no longer splits $GL_K(V)$-equivariantly. This means that any construction which takes a constant coefficient antisymmetric form and produces a primitive for it must divide by $2$ or use an explicit basis (otherwise the formula would still make sense over $K$).

As to your calculation that $\mathrm d\eta = \dot\eta$, the only way I know how to make sense of the left-hand side is to say that $\eta = E$ is the Euler vector field and $\mathrm d$ is the canonical affine connection on the trivial tangent bundle of $V$. Thus $\mathrm d\eta$ should be a one-form with values in the tangent bundle (it is of course the canonical one-form). We do indeed have $\mathrm d\omega(\eta,\dot\eta) = \omega(\dot\eta,\dot\eta)$, but this is not the constant coefficient two-form determined by $\omega$: It takes two tangent vectors and first produces $\dot\eta\wedge\dot\eta(X,Y) = X\otimes Y - Y\otimes X$, then evaluates $\omega$ on it. Since $\omega$ was already antisymmetric, we get the same result twice, which explains the overall factor of $2$.