Have a look at this link:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/ltrans.html#c2
I will use the notation used in the hyperphysics link, as Einstein seems to use notation that is no longer generally used. For example he uses \beta to refer to 1/\sqrt{(1-v^2/c^2)} whereas \beta is more commonly used to refer to v/c these days and \gamma more commonly refers to 1/\sqrt{(1-v^2/c^2)}. In keeping with the commonly used notation, I will call the stationary frame k and the moving frame k'.
Now let us say that all clocks in k have been synchronised with each other in k and likewise all clocks in k' have been synchronised with each other. If we consider ourselves to be frame k the first thing we would notice is that the clocks in k' do not appear to synchronised from our point of view and some are ahead of our clocks and some are behind. Now hyperphysics states that the reference frames coincide at t=t'=0. The first problem is that because all the clocks in k' appear to be showing different times to us, we would be wise to pick one that shows the same time as the clocks in our frame and call that time zero in both frames. The two clocks that are alongside each other momentarily in the two frames and showing the same time give a convenient location for the origins of the two frames so that x.y.z.t = x',y',z',t' = 0. Now if we uses the transformation equation given in the link , e.g:
t = \gamma \left(t' + \frac{vx'}{c^2}\right)
it can be seen that there is a dependency or assumption of zeroing the two clocks when the origins of the two frames coincide. However, if we are only concerned with differences between two events using the equations shown here:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html#c2 e.g:
(t_2 - t_1) = \gamma\left(t_2 ' + \frac{vx_2 '}{c^2}- t_1 ' - \frac{vx_1 '}{c^2}}\right)
then there is no requirement to have the two clocks at the origins of the two frames showing the same time when the origins coincide.
As for the statement "To any time of the stationary system K there then will correspond a definite position of the axes of the moving system" it is self evident that if the moving frame is moving at a constant velocity there will be a correlation between the relative position of the origins of the two frames and the time in the stationary system. Whether or not there is an offset in the times of the origin clocks when the origins of the spatial axes of the two frames coincide, there will still a "definite position" for the origin of the moving frame for any given time in the stationary system. At any given time t in the stationary frame the origin of the moving frame will be at v*t+t_offset. Obviously choosing an t_offset of zero is more convenient, but not an absolute prerequisite.
