A Lagrangian connection and its algebraic interpretation Assume that $M$ is a $n$ dimensional Riemannian manifold. So $TM$ has a natural structure of a symplectic manifold.
A Lagrangian connection $D$ on $M$ is a $n$ dimensional distribution for $TM$   such that for all $z\in TM$, $D_z$ is a lagrangian subspace of $T_z TM$  which is transverse to the vertical foltion of $TM$.

Does every manifold admit a Lagrangian connection?
  Is the $LC$ connection necessarily a Lagranģian connection?

Let $\nabla$ be the corresponding derivation associated with a Lagrangian connection.

What formula is satisfied by $\nabla$?

 A: One possible source of confusion is that $TM$ is not naturally a symplectic manifold in the sense that there is no symplectic structure on $TM$ that is preserved by all diffeomorphisms of $M$.  Meanwhile, $T^*M$ is naturally a symplectic manifold.  (There is no natural identification of $TM$ with $T^*M$ that does not use some extra data, such as a Riemannian metric or a symplectic form on $M$ itself.)  
Maybe you meant to ask this question about connections on $T^*M$?  In that case, the answer is that a connection on $T^*M$ is Lagrangian in your sense if and only if the connection is torsion-free, i.e., the covariant differential $\nabla(\mathrm{d}f)$ (which is a section of $T^*M\otimes T^*M$) is symmetric, i.e., a section of $S^2(T^*M)$, for all smooth functions $f$ on $M$.  More generally, a connection $\nabla$ on $T^*M$ is torsion-free if and only if $\alpha(\nabla \omega\bigr) = \mathrm{d}\omega$ for all $1$-forms $\omega$ on $M$, where $\alpha:T^*M\otimes T^*M\to \Lambda^2(T^*M)$ is the natural anti-symmetrization operation.  (Note that, for me, the 1-form that appears in the covariant differential is a coefficient, i.e., it appears in front of the vector term.  Thus, for example, $\nabla(f\alpha) = \mathrm{d}f\otimes\alpha + f\,\nabla\alpha$.)
In particular, if $\nabla$ is the Levi-Civita connection induced on $T^*M$ by a Riemannian metric $g$, then it is Lagrangian in your sense because it is torsion-free, by definition.
When a Riemannian metric $g$ enters the picture, it causes some confusion because of two things:  First, the symplectic form $\omega_g$ induced on $TM$ by the identification of $TM$ with $T^*M$ via $g$ depends on $g$ in a nontrivial way.
Second, there are now two different ways to define a connection on $TM$ given a connection $\nabla$ on $T^*M$.  The first is by the direct identification of $TM$ with $T^*M$, which defines a connection $\nabla'$ on $TM$ by the formula
$$
\nabla'X = (\mathrm{id}\otimes \sharp) \bigl(\nabla \,\flat(X)\bigr),
$$
where $\flat: TM\to T^*M$ and $\sharp:T^*M\to TM$ are the isomorphisms induced by the metric $g$.  The second is the natural 'dual connection' $\nabla^*$ on $TM$
that satisfies
$$
\mathrm{d}\bigl(\alpha(X)\bigr) = \bigl(\nabla\alpha,X\bigr) + \bigl(\alpha, \nabla^*X\bigr) 
$$
for all $1$-forms $\alpha$ and all vector fields $X$ on $M$ (with the natural pairings assumed).  The answer to your question depends on which one of these ways you use to transfer connections from one vector bundle to the other.  
Assuming that you want to transfer connections by the first method (which, I think, corresponds most closely to the question you asked), then, starting with a connection $\nabla$ on $TM$ and using the 'reverse' formula
$$
\nabla'\alpha = (\mathrm{id}\otimes \flat) \bigl(\sharp(\alpha)\bigr)
$$
to define a connection $\nabla'$ on $T^*M$ (note that this uses the metric $g$), then the condition you seek is that $\nabla'$ be torsion-free, which is the same as saying that the connection $(\nabla')^*$ on $TM$ be torsion-free.  
When you expand this out, you'll see that this condition is equivalent to the following one:  Let $\nabla^g$ be the Levi-Civita connection of $g$.  Then $\nabla-\nabla^g$, which is a tensor, is a section of $T^*M\otimes T^*M\otimes TM$.  Using $g$ to 'lower an index', we get a tensor 
$$
\delta(\nabla,\nabla^g) = (\mathrm{id}\otimes\mathrm{id}\otimes\flat)(\nabla-\nabla^g)
$$
that is a section of $T^*M\otimes T^*M\otimes T^*M$.  Then $\nabla$ is '$g$-Lagrangian' in your sense if and only if $\delta(\nabla,\nabla^g)$ is a section of $T^*M\otimes S^2(T^*M)$.
In particular, since $\delta(\nabla^g,\nabla^g) = 0$, we see that $\nabla^g$ is $g$-Lagrangian in your sense.
