Hydration, Coordinate Bonding, and Reduction Potential in Groups 1 and 2 Elements
This answer addresses your question regarding the reaction

$$\ce{2Li}(s) + \ce{2H2O}(l) \rightarrow \ce{H2}(g) + 2\ce{Li+}(aq) + 2\ce{OH-}(aq)$$


  Whenever an alkali metal such as Lithium is hydrated/solvated,
  coordinate bonding occurs and it is the high hydration enthalpy of
  lithium which makes it the strongest reducing agent in aqueous
  solution. My question is, if lithium is hydrated and becomes
  surrounded by water molecules in coordinate bonding, then how does the
  overall reaction of adding lithium to water produce LiOH and H2(gas) ?
  Is the coordinated ion an intermediate product?


The reaction mechanism for this reaction involves some radical chemistry:



Lithium hydroxide is soluble in water, and will exist as separate ions as long as the solution is not saturated. As soon as the ions are generated, they will be solvated by nearby water molecules. The solvated ions are not intermediates - they might better be viewed as products.

The lithium metal is very reactive due to its free electron, and will react quickly. Besides, Li metal is neutral, and will not immediately form any strong interactions with the polar water molecules, I would think.


  And if Lithium is the strongest reducing agent in aqueous solution I
  assume it means it is reducing water itself, in order to create
  lithium hydroxide and hydrogen gas, but again, how does that come
  about from the coordinate bonding and the hydrated lithium? Coordinate
  bonds preserve the charge of the cation but I am not quite sure how
  this is so.


Indeed, in the course of the reaction water is reduced and Li is oxidized. But I don't think the Li metal (neutral atomic Li) is solvated by water. The reaction with water takes place extremely quickly. And as mentioned above, it is not intuitive to me why water would want to solvate a Li atom in the first place, due to it being non-polar.