How do we deduce that it is the zero function? Theorem
If the Wronskian of $x^{(1)}(t), \dots, x^{(n)}(t)$, that are solutions of $x'=Ax$ on an interval, gets zero at some point $t=t_0$ of the interval, then $x^{(1)}, \dots, x^{(n)}$ are linearly dependent on this interval.
Proof
Let $W(t_0)=0$. Then the vectors $x^{(1)}(t_0), \dots, x^{(n)}(t_0)$ are linearly indepedent, so there are constants $c_1^{\star}, \dots, c_n^{\star}$, at least one of which is non-zero, such that $\sum_{k=1}^n c_k^{\star} x^{(k)}(t_0)=0$. 
We define $x^{\star}(t)= \sum_{k=1}^n c_k^{\star} x^{(k)}(t) (1)$.
$(1)$ is a solution of the system $\frac{dx}{dt}=Ax$ and at $t=t_0, x^{\star}(t_0)=0$. From the theorem of uniqueness, we deduce that there is only one function that satisfies the initial condition $x^{\star}(t_0)=0$.
It is obvious that this function is the zero function, so $\sum_{k=1}^n c_k^{\star} x^{(k)}(t)=0, t \in [a,b]$.
I haven't understood how we deduce that $\sum_{k=1}^n c_k^{\star} x^{(k)}(t)$ is the zero function.
We can't say that it has to hold  $\sum_{k=1}^n c_k^{\star} x^{(k)}(t)=\sum_{k=1}^n c_k^{\star} x^{(k)}(t_0)$ because the initial conditions are the same, since the latter is a function for a specifi value of $t$.
Or am I wrong?

The proof is invoking the uniqueness theorem for solutions to a Cauchy problem. We have the Cauchy problem
$$y'(t) = Ay(t)$$
$$y(t_0) = 0$$
Now, we know there's only one solution to this. But clearly, the constant zero function is a solution, so therefore it's the only solution. Since your function $\sum c^* x^k$ is also a solution, by uniqueness, it equals zero.
This "two functions solve the same Cauchy problem, therefore they're equal" idea is a very common technique in ODEs.
