Given a regular (constant rank) bi-vector $\Pi \in \Gamma(\bigwedge^2TM)$ on a smooth manifold $M$ the necessary and sufficient condition for the image of $\Pi^\sharp:T^\*M\to TM$ to be an integrable distribution is that $\Pi$ is a twisted Poisson tensor with respect to a closed 3-form $\phi$, i.e. there exist a closed 3-form $\phi$ such that
$$[\Pi,\Pi]=\Pi^\sharp(\phi)$$
(see e.g. arXiv:1104.0880).


What is the corresponding condition for a degenerate regular symmetric (2,0)-tensor (a degenerate metric) $g\in \Gamma(S^2TM)$ to produce an integrable distribution as the image of $g^\sharp:T^\*M\to TM$?


And what about any regular (constant rank) tensor $T\in \Gamma(TM\otimes TM)$, with no special symmetry properties?



twisted Poisson structures, degenerate metrics and integrability properties of (2,0)-tensors
Perhaps you will find this a useful answer, at least for the symmetric case: Let $S^2\_r(TM)\subset S^2(TM)$ denote the subbundle consisting of the cometrics of rank $r$. (This bundle is the disjoint union of $r{+}1$ smooth subbundles that are distinguished by the algebraic type of the symmetric bivector. However, this 'type division' won't play a role in this analysis.) 


There is a natural first-order differential operator
$$
\delta\_r:C^\infty\bigl(S^2\_r(TM)\bigr)\longrightarrow 
C^\infty\bigl(\Lambda^2(TM)\otimes \Lambda^r(TM)\otimes \Lambda^{r+1}(TM)\bigr)
$$
with the property that $\delta\_r(g)=0$ if and only if the $r$-rank subbundle $g^\sharp(T^\ast M)\subset TM$ is integrable. 


This differential operator is defined as follows: Locally, $g$ can be written in the form
$$
g = \sum\_{i,j=1}^rg^{ij}\ X\_iX\_j\ ,
$$
where $X\_1,\ldots, X\_r$ are linearly independent vector fields, $g^{ij}=g^{ji}$ are functions, and $\Delta = \det (g^{ij})$ is nonvanishing. Now define (using the summation convention)
$$
\delta\_r(g) = \Delta\ g^{ik}g^{jl}\ \left(X\_i{\wedge}X\_j\ \otimes\ X\_1{\wedge}X\_2{\wedge}\cdots{\wedge}X\_r 
\ \otimes\ [X\_k,X\_l]{\wedge}X\_1{\wedge}X\_2{\wedge}\cdots{\wedge}X\_r\right).
$$
It is easy to verify that this is independent of the choice of basis $X\_i$, and so it is globally defined. It is also obvious that it vanishes if and only if the bundle $g^\sharp(T^\ast M)$ (which is spanned locally by the $X\_i$) is integrable.


Note that $\delta\_r$ vanishes identically unless $2 \leq r < n$, as would be expected. Also, $\delta\_r(\lambda g) = \lambda^{r+2}\ \delta\_r(g)$ for all nonvanishing functions $\lambda$.


If I had considered *oriented* cometrics of rank $r$ (i.e., I had, in addition, fixed an orientation of $g^\sharp(T^\ast M)$), then I could then have got rid of the $\Lambda^r(TM)$ factor in the definition of $\delta\_r$. However, I didn't want to impose the orientability condition, and the $\Lambda^r(TM)$ factor fixes that problem. 


**The almost Poisson case:** Although you already have a criterion in the almost Poisson case, note that there is a similar construction to test the integrability of the bundle $\Pi^\sharp(T^\ast M)$ in the case of an almost Poisson structure $\Pi$ of constant half-rank $r$: Let $\Lambda^2\_r(TM)\subset \Lambda^2(TM)$ be the subbundle of bivectors of half-rank $r$, i.e., its sections are those $\Pi$ such that $\Pi^r$ is nonvanishing as a section of $\Lambda^{2r}(TM)$ while $\Pi^{r+1}\equiv0$. 


There is a natural first-order differential operator 
$$
\partial\_r:C^\infty\bigl(\Lambda^2\_r(TM)\bigr)\longrightarrow 
 C^\infty\bigl(\Lambda^2(TM)\otimes\Lambda^{2r+1}(TM)\bigr)
$$
defined as follows: Locally, write $\Pi$ in the form
$$
\Pi = \sum\_{i,j=1}^{2r} {\tfrac12}\ a^{ij}\ X\_i\wedge X\_j\ ,
$$
where $X\_1,\ldots,X\_{2r}$ are linearly independent vector fields, $a^{ij}=-a^{ji}$ are functions, and $\mathrm{Pf}(a^{ij})$ is nonvanishing (since $\Pi^r = r!\ \mathrm{Pf}(a^{ij})\ X\_1\wedge X\_2\wedge\cdots\wedge X\_{2r}\not=0$). Now define (again using the summation convention)
$$
\partial\_r(\Pi) = 
 {\tfrac14}\ a^{ik}a^{jl}\ 
 X\_k{\wedge}X\_l\otimes\ [X\_i, X\_j]{\wedge}\Pi^r\ .
$$
It is easy to verify that this does not depend on the choice of basis $X\_i$, 
and so it is well-defined. 


It is evident that that $\partial\_r(\Pi)$ vanishes if and only if the bundle $\Pi^\sharp(T^\ast M)$ (which is spanned locally by the $X\_i$) is integrable. Also, $\partial\_r(\lambda\ \Pi)= \lambda^{r+2}\partial\_r(\Pi)$ for all nonvanishing functions $\lambda$.


This is a somewhat more explicit test that the one you proposed, I think.


**The general case:** It turns out that one does not need any assumption of symmetry or skew-symmetry to get the integrability tensor, just an assumption of constant rank. Let $\otimes^2\_r(TM)\subset TM\otimes TM$ denote the subbundle consisting of the tensors of rank $r$. A section $\phi\in C^\infty\bigl(\otimes^2\_r(TM)\bigr)$ can be written locally in the form 
$$
\phi = \sum\_{i,j=1}^r f^{ij}\ X\_i\otimes Y\_j
$$
where $X\_1,\ldots,X\_r$ are linearly independent vector fields, $Y\_1,\ldots,Y\_r$ are linearly independent vector fields, and $f=(f^{ij})$ is an invertible matrix, i.e., $\det(f)\not=0$. The vector fields $X\_i$ are local sections of a globally defined bundle $\lambda\_\phi\subset TM$ of rank $r$, and the vector fields $Y\_i$ are local sections of a globally defined bundle $\rho\_\phi\subset TM$ of rank $r$.


There is a natural first-order differential operator
$$
D\_r: C^\infty\bigl(\otimes^2\_r(TM)\bigr)\longrightarrow
 C^\infty\bigl(\Lambda^2(TM)\otimes \Lambda^r(TM)\otimes \Lambda^{r+1}(TM)\bigr)
$$
which, relative to the local expression given above, takes the form
$$
D\_r(\phi) = 
\det(f)\ f^{ik}f^{jl}\ \left(Y\_k{\wedge}Y\_l\ \otimes
\ Y\_1{\wedge}Y\_2{\wedge}\cdots{\wedge}Y\_r 
\ \otimes\ [X\_i,X\_j]{\wedge}X\_1{\wedge}X\_2{\wedge}\cdots{\wedge}X\_r\right).
$$ 
This operator has the property that $D\_r(\phi)=0$ if and only if $\lambda\_\phi$ is integrable. 


Note that there is a well-defined involution $\iota$ on $C^\infty\bigl(\otimes^2\_r(TM)\bigr)$ such that
$$
\iota(\phi) = \sum\_{i,j=1}^r f^{ij}\ Y\_j\otimes X\_i\ ,
$$
and the integrability of $\rho\_\phi$ is equivalent to the equation $D\_r\bigl(\iota(\phi)\bigr)=0$.


