Theorem
Let $x_i^{(k)}(t), i, k=1, \dots, n$ be a fundamental system of solutions of $x'=Ax$. Then any solution of this system can be written as a linear combination of $x_i^{(k)}(t), i,k=1, \dots, n$ with suitable constants, i.e. in the form
$$x(t)= \sum_{k=1}^n c_k x^{(k)}(t)$$
Proof
Let $x(t)$ be a solution of $\frac{dx}{dt}=Ax$, where $x(t_0)=c_0$ (i.e. $(x_1(t_0), \dots, x_n(t_0))=(c_{01}, \dots, c_{0n})$).
Since $W(t_0) \neq 0$ we have that $x^{(k)}(t_0)$, $k=1, \dots, n$ are linearly independent vectors, thus there are such constants $c_k^{\star}, k=1, \dots, n$ that $c_0= \sum_{k=1}^n c_k^{\star} x^{(k)}(t_0)$.
We define $x^{\star}= \sum_{k=1}^n c_k^{\star} x^{(k)}(t)$.
$x^{\star}(t)$ is a linear combination of solutions of the system $\frac{dx}{dt}=Ax$, so it is a solution of this system. Because of construction, $x^{\star}(t_0)=c_0$ so from the theorem of uniqueness it follows immediately that $x^{\star}(t) \equiv x(t)$, i.e.
$$x(t)= \sum_{k=1}^n c_k^{\star} x^{(k)}(t)$$
At this part: "Since $W(t_0) \neq 0$ we have that $x^{(k)}(t_0)$, $k=1, \dots, n$ are linearly independent vectors, thus there are such constants $c_k^{\star}, k=1, \dots, n$ that $c_0= \sum_{k=1}^n c_k^{\star} x^{(k)}(t_0)$."
are we talking about the Wronskian of the solutions $x_i^{(k)}(t)$ of the fundamental system of solutions? If so , doesn't it hold that $W(t) \neq 0$ for any $t$ ?
Also why do we deduce that $c_0= \sum_{k=1}^n c_k^{\star} x^{(k)}(t_0)$ and not $0= \sum_{k=1}^n c_k^{\star} x^{(k)}(t_0)$ ?