The symplectic area contained in a closed curved, that is the boundary of map of a disc, is the "action along the curve". 
$$
\int_\sigma \omega = \int_\sigma d\lambda = \int_{\partial \sigma} \lambda = \int_0^{2\pi} \lambda_{\gamma(t)}(\dot \gamma(t)) dt,
$$
where $\sigma$ is a smooth map from the disc to $M$, and $\gamma = \partial \sigma$. In all cases, the pullback of the 2-form $\omega$ by $\sigma$ is exact since the disc is contractible, so there exists a primitive $\lambda$, on the disc, and you apply Stokes' theorem.

---

Let me try to elaborate a little bit on a not too complicate but not that simple example to see where the symplectic form makes sense. Let us consider a point on the sphere $S^2$, let 
$$
TS^2 = \{ (x,v) \in S^2 \times {\bf R}^3 \mid x \cdot v = 0 \}
$$
Let 
$$
L : TS^2 - S^2 \to {\bf R} \quad \mbox{with} \quad L(x,v) = \Vert v \Vert
$$
be the "length function" as lagrangian. And you look for the variational problem
$$
 \delta \int L(x(t),\dot x(t))\ dt = \delta \int \Vert \dot x(t) \Vert\ dt = 0.
$$
I don't put the limits of the integral on purpose, it would lead to a too long discussion. Since the lagrangian is homogeneous of degree 1 in $v$, we have the Euler identity
$$
L(x,v) = \frac{\partial L(x,v)}{\partial v}(v)
$$
And the nature of the partial derivative involved above is a map from $TS^2-S^2$ to the cotangent $T^*S^2$
$$
\forall v \in T_xS^2 - \{0\}, \quad \frac{\partial L(x,v)}{\partial v} = \frac{\bar v}{\Vert v \Vert} \in T^*_xS^2
$$
where the bar denotes the transposed. Let's call this map $P$
$$
P : TS^2 - S^2 \to T^*S^2 \quad \mbox{with} \quad P(x,v) = \frac{\partial L(x,v)}{\partial v}.
$$
Now let $\lambda = pdx$ the Liouville form on $T^*S^2$, its pullback by $P$, integrated along the curve $\gamma = [t \mapsto (x(t),\dot x(t))]$ is exactly the action
$$
\int \Vert \dot x(t) \Vert \ dt = \int_\gamma P^*(\lambda) = \int_{P \circ \gamma} \lambda.
$$

[[To be continued]]