Intuition for the Lax-Milgram Theorem

Regarding the first question:

This $u$ is what we call a weak solution. The most obvious reason why we ask for weak solutions is because sometimes strong solutions don't exist and we still want to make sense of problems in these cases. The other reason that comes to mind is that even when strong solutions exist, it is still often easier to prove that by showing existence of a weak solution and then a posteriori showing it is a strong solution. This is in part because weak solutions provide a framework for the use of variational methods and so forth.

Regarding the third question:

The finite dimensional analogue of Lax-Milgram is that $Ax=b$ has a unique solution if $A$ is symmetric positive definite. The "weak formulation" of $Ax=b$ is $\langle v,Ax \rangle = \langle v,b \rangle$ for all $v \in \mathbb{R}^n$, and the bilinear form $a$ now is $a(v,x)=\langle v,Ax \rangle$. (I use scare quotes on weak formulation because the two formulations are equivalent.)

Regarding the second question:

Relative to the finite dimensional analogy described in the previous paragraph, you are converting from $Ax=b$ to $x=\arg \min f$ where $f(x)=\frac{1}{2} \langle x,Ax \rangle - \langle x, b\rangle$. The fact that $x=\arg \min f$ implies $Ax=b$ follows from writing out the equation $\nabla f = 0$ and applying the symmetry of $A$. The positive definiteness is what you need to ensure that the resulting critical point is a minimum.