Note: $S$ and $S'$ are just frames given in the figure, where $S$ is an inertial frame ( it is not accelerating ) .
From this proof of finding pseudo force , we can see that $m a_0$ is a pseudo force here. Now , I see pseudo force here is mass multiplying the acceleration of $S'$ with respect to S.
So , is it like Force on the $S'$ with respect to S. Then , what is value $m$ for ?
When you are in a non inertial (accelerating) reference frame you see things you are not supposed to see. In those frames, $F=ma$ does not hold.
Imagine you are on a train. You jump and while you are in the air the train decelerates abruptly. Because when you left the floor you where moving with the train's speed, you keep moving at that speed during your jump whereas the train stops under you and you fall down at a different position than the one you started with despite jumping vertically. If you try to analyse this motion using $F=ma$ you fail: it simply does not work (you jumped in the vertical direction and moved in the horizontal one?!). What was the force that pushed you? The answer is that no force pushed you: your movement was an effect to a change in frame of reference!
However if you "invent" a force given by $F_i=-mA_i$ where $A_i$ is the train acceleration and $m$ the mass of the object you are considering (you, in this case) you recover a full description of your system. The demonstration is in the page of the book you attached. Basically, you transform the "extra" acceleration that you feel because of the frame of reference being non inertial into a "fake" (pseudo) force which obeys Newton's law $F=ma$. You moved as if a force $F_i=-mA_i$ was acting on your body.
So pseudo-forces are not "real" forces in a common sense: they are frame-dependent, they can be measured (but what one is actually measuring is the acceleration of the frame, not a force in the strict sense..), they have a clear effect but they don't derive from any field, any charge, any gravitational mass, etc! They are somewhat a mathematical artifact adopted to make sure $F=ma$ holds in every frame of reference and but at the same time... they act on objects.
