Daniel, I understand what you mean. Sometimes it's difficult to understand abstract proofs, especially when the proofs only involve many variables/symbols/greek letters and not numbers.
What usually works for me is to get 1 or 2 examples (either from books, online, lectures, etc.) which illustrate how the theorem works. I feel that this "concreteness" helps me understand what is going on better, and then I go back and re-read the proof the of theorem. Usually easy, trivial examples are best.
Then, when I've understood the theorem and its proof well, I can try to apply it to more difficult examples.
So for example in Linear Algebra, when learning the Rank-Nullity Theorem, you could find a very simple example (maybe 2x2 matrices) from a textbook first to see it in practice. And perhaps try another yourself to see the Rank-Nullity Theorem in practice. Then afterwards go back and re-read the proof.
Hope that helps.
The determinant formula isn't so mysterious. Consider the cross product $\mathbf{v} = \langle a,b,c \rangle \times \langle d,e,f \rangle$ as the formal determinant
$$ \det \left(\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ a & b & c \\ d & e & f \end{array} \right) $$
where $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are the standard basis vectors. If instead one considers $\mathbf{i}, \mathbf{j}, \mathbf{k}$ as indeterminates and substitutes $x, y, z$ for them, this determinant computes the dot product $\mathbf{v} \cdot \langle x, y, z \rangle$. But letting $\langle x, y, z \rangle$ be $\langle a, b, c \rangle$ or $\langle d, e, f \rangle$ gives a zero determinant, so $\mathbf{v}$ is perpendicular to the latter two vectors, hence to the plane they span, as Omnomnomnom says.
Best Answer
Geometrically Speaking:
The dot product is vital for measuring lengths and angles. First ${\bf u} \cdot {\bf u} = ||{\bf u}||^2$ and second, assuming ${\bf u}$ and ${\bf v}$ are non-zero, we have ${\bf u} \cdot {\bf v} = 0$ if and only if ${\bf u}$ and ${\bf v}$ are perpendicular.
The cross product of two non-zero, non-parallel vectors gives a third vector that is perpendicular to the first two. Thus: ${\bf u} \perp ({\bf u} \times {\bf v}) \perp {\bf v}.$
Moreover, the dot product and the cross product are related by the triple scalar product:
$$[{\bf u},{\bf v},{\bf w}] = ({\bf u} \times {\bf v})\cdot {\bf w}$$
The triple scalar product is the volume of the parallelepiped spanned by ${\bf u}$, ${\bf v}$ and ${\bf w}$.
Abstractly Speaking:
The dot product can be used to show that a vector space $V$ is isomorphic to its dual vector space $V^*$. Recall that $V^*$ consists of all linear maps $f : V \to \mathbb{R}.$ The isomorphism $\phi : V \to V^*$ is given by $\phi : {\bf u} \mapsto f_{{\bf u}}$ where $f_{{\bf u}}({\bf v}) = {\bf u} \cdot {\bf v}$ for all ${\bf v}$ in $V$.