Let $p:X \to Y$ be a covering and $f_0,f_1$ two pathhomotopic paths in $Y$.Let $f_0'$, $f_1'$ be two liftings with $f_0'(0)=f_1'(0)$. Then $f_0'$, $f_1'$ are homotopic and consequently $f_0'(0)=f_1'(0)$.
The proof should be a corrollary of the Covering Homotopy Theorem. But I dont understand one point. Let $F$ be a homotopy between $f_1,f_2$. Then as $f_1'$ is by definition equal to $F$ on $[0,1]\times \{0\}$. By homotopy lifting Theorem there is a unique lifting $G:[0,1]^2 \to X$ such that the diagram comutes (Im not able to draw it hier without tikz) i.e $G$ equals $f_0'$ on $[0,1]\times\{0\}$ and $pG=F$. Now we claim that $G$ is a homotopy between $f_0'$, $f_1'$. In fact $G(x,0)= f_0'(x)$ and $G$ is constant on the boundary of $I$ (by homotopy Theorem). The problem is now $G(x,1)= f_1'(x)$. Why is that true? I tried the following:
Let $G(x,1) = h(x)$. Then By the commutativity of the diagram: $f_1(x) = F(x,1)=p(G(x,1))=p(h(x))$. So $h$ is a a lifting of $f_1$. I would like to use that liftings of paths are unique but this is only true at least to my knowledge if both liftings start at the same point i.e I would need $h(0)= f_1'(0) =f_0'(0)$. Is that true and if yes why?
You have the right approach, but I think it should be elaborated a bit better. Here is my suggestion.
Let $I = [0,1]$. For each homotopy $H : A \times I \to B$ we write $H_t : A \to B$ for the map defined by $H_t(a) = H(a,t)$.
Let $y_0 = f_0(0) = f_1(0)$ be the common initial point and $y_1 = f_0(1) = f_1(1)$ the common terminal point of the paths $f_0, f_1$.
Let $F : I \times I \to Y$ be a homotopy from $f_0$ to $f_1$ which is stationary at the points $0, 1 \in I$. There exists a unique lift $G : I \times I \to Y$ such that $G_0 = f'_0$. The homotopy $G$ is stationary at the points $i = 0,1 \in I$ because $G(\{i\} \times I)$ is a connected subset of the fiber $p^{-1}(y_i)$. Since all fibers are discrete, $G(\{i\} \times I)$ must be a singleton.
Hence the map $G_1$ is a lift of $f_1$ which has the same initial and terminal points as $f'_0$. But we know that $f'_1$ is a lift of $f_1$ having the same initial point as $f'_0$. By unique path lifting we see that $G_1 = f'_1$.