Why are forces independent from the frame of reference? The following question occurred to me while reading a proof of the following statement:

If K is an inertial frame of reference, then a K’ frame of reference,
  which is moving with a constant velocity compared to K, is also an
  inertial frame of reference. 

In the proof $\sum F=\sum F'$ is used and I'd like to know why is this equation valid.  

Here goes the proof:  

Let $\underline{r}(t)$ be the position vector of the point $P$ in the
  $K$ frame.   Let $\underline{r'}(t)$ be the position vector of the
  point $P$ in the $K'$ frame.   Let $\underline{r}_{K'}(t)$ be the
  position vector of the point $K'$ in the $K$ frame.  

 

$K’$ is moving with a constant velocity compared to $K$, therefore:
  $$\begin{equation} \underline{r}_{K'}(t)=\underline{w}\cdot
 t+\underline{r_0} \end{equation}$$ where $\underline{w}$ is the
  velocity vector of $K'$ and $\underline{r_0}$ is the position vector
  of the origo of $K'$ (in the $K$ reference frame) in the $t=0$ moment.
  The connection between the two position vectors: $$
\underline{r}(t)=\underline{r'}(t)+\underline{r}_{K'}(t)=\underline{r'}(t)+\underline{w}\cdot t+\underline{r_0}$$ This is the Galileian-transformation.   After
  derivating with respect to time twice we get: $$ \begin{aligned}
 \underline{v}(t) &=\underline{v'}(t)+\underline{w} \\ Eq. 1:
 \underline{a}(t) &=\underline{a'}(t) \end{aligned} $$ Newton II. in
  the inertial frame of reference $K$ $$ m
 \underline{a}=\sum\underline{F} $$ where $\sum\underline{F}$ is the
  net force in the $K$ frame.   Combining this formulae with $Eq. 1$ and
  using that  $\sum\underline{F}=\sum\underline{F}'$ ( forces are
  independent from the frame of reference) we get: $$ \sum
 \underline{F}'=m\underline{a'}$$ which implies that $K'$ is an
  inertial reference frame. $ \blacksquare $

 A: If an object has no acceleration in one inertial frame of reference that means no real forces acting on it.
  now suppose you  observe the same object from a different inertial frame,its not possible that just because you are observing the same object from a different inertial frame somehow a real force will start acting on the object.
But if you observe now the same object from an noninertial frame of reference then fictitious forces will act on the object and here the object can have acceleration without some real forces acting on it.
That means you first calculate the net force acting on a particle in one inertial frame of reference and then calculate net force from a different frame of reference(Inertial or noninertial) .
If the net force is not same(for both frames) means some fictitious forces are acting on the particle.
If there are fictitious forces then you are observing now from a non inertial frame of reference.
A: If the forces are determined, say, by the relative position of two objects, then the relative position of the two objects is the same in the two frames.  If the force is determined, say, by the relative velocity of two objects, then the relative velocity of the two objects is the same in the two frames.
Many forces are like this; a spring, friction due to air resistance, Newton's gravitational force, etc.
A: Suppose an object is experiencing a certain total force $\mathbf F$ applied to it. If you change to a different frame of reference which is inertial w.r.t. to the previous one (i.e. they are related by a Galilei transformation, that is to say one is moving with constant velocity w.r.t. the other), then there is no apparent acceleration added to the object, since $\mathbf a = \ddot{\mathbf x}$, and the relation between coordinates in the two frame of reference are at most linear in $t$, whence the extra acceleration is $\mathbf a=0$. Therefore forces, whether real or apparent, are conserved when passing to a different frame of reference through a Galilei transformation.
