Let me address why the Wronskian works. To begin let's use vectors of functions (not necessarily solutions of some ODE).
For convenience, I'll just work with $3 \times 3$ systems.
Let $$ {\bf f}_1(t) = \begin{bmatrix} f_{11}(t) \\ f_{21}(t) \\ f_{31}(t) \end{bmatrix}, \qquad {\bf f}_2(t) = \begin{bmatrix} f_{12}(t) \\ f_{22}(t) \\ f_{32}(t) \end{bmatrix},
\qquad \mathrm{and}
\qquad {\bf f}_3(t) = \begin{bmatrix} f_{13}(t) \\ f_{23}(t) \\ f_{33}(t) \end{bmatrix} $$
be vectors of functions (i.e. functions from $\mathbb{R}$ to $\mathbb{R}^3$).
We say the set $\{ {\bf f}_1(t), {\bf f}_2(t), {\bf f}_3(t) \}$ is linearly dependent on $I \subseteq \mathbb{R}$ (some set of real numbers) if there exist $c_1,c_2,c_3 \mathbb{R}$ (not all zero) such that $c_1{\bf f}_1(t)+c_2{\bf f}_2(t)+c_3{\bf f}_3(t)={\bf 0}$ for all $t \in I$. [Be careful here: this equation must hold for all $t$'s in $I$ simultaneously with the same constants.]
This equation can be recast in terms of matrices. We have linear dependence if and only if
there exists some constant vector ${\bf c} \not= {\bf 0}$ such that ${\bf F}(t){\bf c}={\bf 0}$ for all $t \in I$. This is where ${\bf F}(t) = [{\bf f}_1(t) \;{\bf f}_2(t) \;{\bf f}_3(t)]$. Or writing it out in a more expanded form:
$$ {\bf F}(t){\bf c} = \begin{bmatrix} f_{11}(t) & f_{12}(t) & f_{13}(t) \\ f_{21}(t) & f_{22}(t) & f_{23}(t) \\ f_{31}(t) & f_{32}(t) & f_{33}(t) \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$
Now the determinant of ${\bf F}(t)$ is known as the Wronskian of the functions ${\bf f}_1,{\bf f}_2,{\bf f}_3$. That is $W(t) = \mathrm{det}({\bf F})(t)$.
Now we call on basic Linear algebra. The columns of an $n \times n$ matrix $A$ are linearly dependent if and only if there is a non-trivial (i.e. non-zero) solution of $A{\bf x}={\bf 0}$. This is true if and only if $\mathrm{det}(A)=0$.
But be very careful this is for a system of constants (not functions). To show the columns of ${\bf F}$ are linearly dependent we need there to be a non-zero solution for all $t$ in $I$.
So only the following can be said: IF the columns of ${\bf F}(t)$ are linearly dependent on $I$, THEN there is a non-zero solution for ${\bf F}(t){\bf c}={\bf 0}$ which works for all $t$ in $I$. Thus $W(t)=\mathrm{det}({\bf F})(t)=0$ for all $t$ in $I$.
The converse does not hold in general.
However, for sets of solutions of linear systems of ODEs, Abel's Identity shows that the Wronskian is a constant multiple of an exponential function. Thus if it's zero somewhere, it's zero everywhere (well, everywhere these are solutions anyway). So in intro to DEs classes, professors will commonly state that we have linear dependence if and only if the Wronskian zero and then proceed to use examples for which this theorem isn't necessarily true! The implication does not go both ways in general.
Now finally, how to connect this back to regular functions? Well, consider functions $f,g,h$. Then they are linearly dependent on some set of real numbers $I$ if we can find $a,b,c \in \mathbb{R}$ (not all zero) such that $af(t)+bg(t)+ch(t)=0$ for all $t$ in $I$.
If we differentiate this again and again, we'll get other equations:
$$ \begin{array}{ccc} af(t)+bg(t)+ch(t) & = & 0 \\ af'(t)+bg'(t)+ch'(t) & = & 0 \\ af''(t)+bg''(t)+ch''(t) & = & 0 \end{array} $$
Now we're back to the discussion about linear independence in reference to the set:
$$ \left\{ \begin{bmatrix} f(t) \\ f'(t) \\ f''(t) \end{bmatrix}, \begin{bmatrix} g(t) \\ g'(t) \\ g''(t) \end{bmatrix}, \begin{bmatrix} h(t) \\ h'(t) \\ h''(t) \end{bmatrix} \right\} $$
So the Wronskian you're using is a special case of the Wronskian for systems of ODEs.
I hope this clears up a little!
