Suppose I am given a $2n$ dimensional manifold $M$ and I want to put a symplectic form on it. How could I determine whether or not it admits one and if I do know that it admits one what are my options for (explicitly) constructing some?
My current knowledge of techniques is the following:
(1) If we are in $\mathbb R^{2n}$ or $\mathbb C^n$ then we can take global coordinates and construct the standard ones.
(2) If we are on a tangent bundle we can construct the canonical one using the tautological 1-form.
(3) If the Riemannian holonomy (assuming we put some metric on $M$ -- although I'm confused about why the metric we choose matters...) group is contained in the symplectic group then we can transfer symplectic forms on $\mathbb R^{2n}$ to $M$. My understanding here though is that we can say the form looks a certain way in certain coordinates at a point on the manifold but not really what it looks like locally.
(4) Brute force- define locally in coordinates and then show well-defined.
(5) And I am aware of this cohomology existence result.
So are there any other techniques? In particular, ways like method $(2)$, where one can construct the form globally but still know what it looks like locally.
In your list of constructions, you are certainly missing coadjoint orbits: If $G$ is any Lie group with Lie algebra $\mathfrak g$, let $\mathfrak g^*$ be the dual vector space to $\mathfrak g$. This carries a natural representation of $G$, defined by $(g\cdot\lambda)(X):=\lambda(Ad(g^{-1})(X))$ with $Ad$ denoting the usual adjoint action. Any orbit of $G$ in this representation is a smooth submanifold (since it is a homogeneous space of $G$) which inherits a canonical symplectic structure, usually called the Kirillov-Kostant-Souriau form.
However, I think that overall your question is rather misleading: By Darboux's theorem, all symplectic forms look the same locally, i.e. locally around any point, you can find coordinates in which a given symplectic form looks like the standard form on $\mathbb R^{2n}$. Global existence of symplectic forms is a very difficult question, and apart from the obvious topological obstructions that occur in your question (and which basically only concern compact manifolds) non-existence results depend on subtle invariants.
By the way, the condition in (3) is a typical example for why I consider your question as misleading. The condition on holonomy says that the given metric $g$ is a Kähler metric. This implies that its so-called fundamental two-form is a symplectic form. Hoever, Kähler metrics are much more restrictive objects than symplectic forms, and even if a manifold admits one Kähler metric, then there is an infinite dimensional space of metrics on $M$ which are not Kähler. So this is a method which only works if you have guessed the right result from the beginning. This does not mean that the condition is not interesting, but you get the existence of a symplectic form from existence of a much stronger and much more rigid structure.