Luboš Motl already answered your subquestions very well, but I thought I'd work out an example in detail for your understanding.
Some people like to work with fictitious forces, most people don't. The thing is that Newton's laws don't work in reference frames that have a non-zero (finite) acceleration. In those reference frames we need to resort to mysterious fictitious forces to make things add up. In inertial frames (frames with zero acceleration), Newton's laws work fine and we need no hocus-pocus.
Now for my example, let's look at the situation of a satellite of mass $m$ orbiting the earth. In the reference frame $O'$ of the satellite, there's only one force exerted on it: the gravitational pull of the earth. In its own reference frame, the satellite is also stationary, so its acceleration with respect to $O'$ is zero. Now, we know this can't be the whole story, otherwise Newton's second law $m\mathbf{a} = \mathbf{F}$ would yield
$$\mathbf{a} = G\frac{M}{r^2}\mathbf{\hat{r}'}$$
where $G$ is the gravitational constant, $M$ is the mass of the earth and $\mathbf{r'}$ is the vector from the satellite to the earth's center. This would mean the acceleration is non-zero, which we know is not the case. So we have to add a fictitious centrifugal force to get $a$ to be zero:
$$m\mathbf{a} = G\frac{mM}{r'^2}\mathbf{\hat{r}'} + \mathbf{F}_{cf}$$
where the centrifugal force $\mathbf{F}_{cf}$ is equal to $-m\frac{v^2}{r'}\mathbf{\hat{r}'}$ where $v$ is the orbital velocity and to get $\mathbf{a}=\mathbf{0}$ we demand that this force cancels out with gravity (a condition from which we can derive the orbital speed!).
A different way of looking at this is by saying: well, we know the reference frame $O'$ isn't inertial, because the second law of Newton gives us nonsense (this doesn't necessarily mean you're not in an inertial frame but let's assume it does). So let's take the reference frame fixed to the center of the earth $O$, which is a fair approximation of an inertial frame for these purposes. From this reference frame we can clearly observe that the satellite's reference frame $O'$ moves with some finite acceleration $\mathbf{a}_{O'}$. Moreover, since it's moving (approximately) on a circle, we know an expression for this acceleration: $$\mathbf{a}_{O'} = -\frac{v^2}{r}\mathbf{\hat{r}}$$ where $v$ is the orbital velocity and $\mathbf{\hat{r}}$ is the vector from the center of the earth to the satellite. So let's insert this acceleration into the equation:
$$m\mathbf{a}_{O'} = -G\frac{mM}{r^2}\mathbf{\hat{r}}$$
(notice the sign change because $\mathbf{\hat{r}} = -\mathbf{\hat{r}'}$) We can rewrite this as follows:
$$\begin{align}
\mathbf{0} &= -G\frac{mM}{r^2}\mathbf{\hat{r}} - m\mathbf{a}_{O'} \\
\mathbf{0} &= -G\frac{mM}{r^2}\mathbf{\hat{r}} + m\frac{v^2}{r}\mathbf{\hat{r}}
\end{align}$$
This is the same equation as before. So we end up with the same physics, but the handling of the problem is much more natural in inertial frames of reference. We needed some artificial construct to get things right in the non-inertial frame, while we had no issues at all in the inertial frame.
Another way of looking at fictitious forces is with fictitious accelerations: the maths are all the same, but the mindset is different. Instead of adding a fictitious force, which you know isn't real, you add a "fictitious" acceleration, which is more true to the actual physics since it's the acceleration of the reference frame if you were to look at it from an inertial frame. If you add the fictitious acceleration $\mathbf{\hat{r}'}v^2/r'$ in the left hand side of the equation $m\mathbf{a} = \mathbf{F}$ for the reference frame $O'$ attached to the satellite (where $\mathbf{a}$ was zero), you get:
$$m\left(\mathbf{0}+\frac{v^2}{r'}\mathbf{\hat{r}'}\right) = G\frac{mM}{r^2}\mathbf{\hat{r}'}$$
This is again the same equation.
