Suppose that $y_1(t), \ldots, y_n(t)$ are solutions of $\frac{d^n y}{dt} + p_{n-1}(t) \frac{d^{n-1} y}{dt} + \cdots + p_1(t) \frac{dy}{dt} + p_0(t) y = 0$, and suppose that their Wronskian is zero for $t = t_0$, i.e.
\begin{equation*}
\left|
\begin{array}{cccc}
y_1(t_0) & y_2(t_0) & \cdots & y_n(t_0) \\
y_1'(t_0) & y_2'(t_0) & \cdots & y_n'(t_0) \\
\vdots & \vdots & \ddots & \vdots \\
y_1^{(n-1)}(t_0) & y_2^{(n-1)}(t_0) & \cdots & y_n^{(n-1)}(t_0)
\end{array}
\right| = 0.
\end{equation*}
Then the corresponding matrix is not invertible, and the system of equations
\begin{array}{c}
c_1 y_1(t_0) &+& c_2 y_2(t_0) &+& \cdots &+& c_n y_n(t_0) &=& 0 \\
c_1 y_1'(t_0) &+& c_2 y_2'(t_0) &+& \cdots &+& c_n y_n'(t_0) &=& 0 \\
\vdots &+& \vdots &+& \ddots &+& \vdots &=& 0 \\
c_1 y_1^{(n-1)}(t_0) &+& c_2 y_2^{(n-1)}(t_0) &+& \cdots &+& c_n y_n^{(n-1)}(t_0) &=& 0 \\
\end{array}
has a nontrivial solution for $c_1, c_2, \ldots, c_n$ not all zero.
Let $y(t) = c_1 y_1(t) + \cdots + c_n y_n(t)$. Because $y(t)$ is a linear combination of solutions of the differential equation, $y(t)$ is also a solution of the differential equation. Additionally, because the weights satisfy the above system of equations, we have $y(t_0) = y'(t_0) = \cdots = y^{(n-1)}(t_0) = 0$.
These initial conditions and the original differential equation define an initial-value problem, of which $y(t)$ is a solution. If $p_0(t), p_1(t), \ldots, p_{n-1}(t)$ are continuous, then any initial-value problem associated with the differential equation has a unique solution. Obviously $y^*(t) = 0$ is a solution of the initial-value problem; since we know that $y(t)$ is also a solution of the same initial-value problem, it follows that $y(t) = 0$ for all $t$, not just $t = t_0$.
We now have $c_1 y_1(t) + \cdots + c_n y_n(t) = 0$ for all $t$, where $c_1, \ldots, c_n$ are not all zero. Thus the functions $y_1(t), \ldots, y_n(t)$ are linearly dependent.
Conversely, if the functions $y_1(t), \ldots, y_n(t)$ are linearly dependent, then the system of equations
\begin{array}{c}
c_1 y_1(t) &+& c_2 y_2(t) &+& \cdots &+& c_n y_n(t) &=& 0 \\
c_1 y_1'(t) &+& c_2 y_2'(t) &+& \cdots &+& c_n y_n'(t) &=& 0 \\
\vdots &+& \vdots &+& \ddots &+& \vdots &=& 0 \\
c_1 y_1^{(n-1)}(t) &+& c_2 y_2^{(n-1)}(t) &+& \cdots &+& c_n y_n^{(n-1)}(t) &=& 0 \\
\end{array}
has a nontrivial solution for every $t$, the corresponding matrix is not invertible for any $t$, and $W[y_1, \ldots, y_n](t) = 0$.
You can directly use Abel's identity to show that if the Wronskian of any two solutions of the differential equation $y''+p(x)y'+q(x)y = 0$ (on an interval $I$) is constant, then $p(x) = 0$. (If $\int_{x_0}^{x} p(t) dt$ is constant $\forall$ $x \in I$, then $p(t) = 0 $)
The answer to your last question can be found here.
Best Answer
The Wronskian matrix $W(y_1, y_2)$ of two solutions $y_1$ and $y_2$ of
$y'' + p(t)y'(t) + q(t)y(t) = 0 \tag 1$
may be defined as
$W(y_1, y_2) = \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix}, \tag 2$
with determinant
$\Delta_W = \vert W(y_1, y_2) \vert = \det \left (\begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix} \right ) = y_1y_2' - y_2 y_1'; \tag 3$
we calculate
$\Delta_W' = y_1'y_2' + y_1y_2'' - y_2'y_1' - y_2y_1'' = y_1y_2'' - y_2y_1''; \tag 4$
we may now use (1) in the form
$y_i'' = -py_i' - qy_i, \; i = 1, 2, \tag 5$
to transform (4) to
$\Delta_W' = y_1(-py_2' -qy_2) - y_2(-py_1' - qy_1) = -py_1y_2' - qy_1y_2 + py_1'y_2 + qy_1y_2 = -p(y_1y_2' - y_1'y_2) = -p \Delta_W, \tag 6$
a simple first order, linear ordinary differential equation for $\Delta_W$; the solutions of this equation are
$\Delta_W(t) = \Delta_W(t_0) e^{-\int_{t_0}^t p(s)\; ds}; \; t_0, t \in I, \tag 7$
which the reader may easily check. It follows from this equation and the uniquenness of solutions that if
$\Delta_W(t_0) = 0, \tag 8$
then
$\Delta_W(t) = 0, \; \forall t \in I; \tag 9$
thus if
$y_1(t_0) = y_2(t_0) = 0, \tag{10}$
it follows that
$\Delta_W(t_0) = 0, \tag{11}$
and (9) binds as well. Thus $y_1$ and $y_2$ cannot be a fundamental solution system for (1) on $I$ since fundamental systems are characterized by the non-vanishing of $\Delta_W$ everywhere, that is, by the linear independence of the columns of (2); but of course when the columns are linearly dependent then $W(y_1, y_2) = 0$.
Finally, as pointed out by user539887 in his comment to the question itself, it's not that a fundamental system doesn't exist but that $y_1$, $y_2$ does not form one. A fundamental solution system always exists, as may be seen by taking
$W(t_0) = I = \begin{bmatrix} 1& 0 \\ 0 & 1 \end{bmatrix}, \tag{12}$
for then
$\Delta_W(t) \ne 0, \; \forall t \in I, \tag{13}$
as follows from (7) since
$\Delta_W(t_0) = 1 \tag{14}$
in this case.