What does Feynman mean when he says that $F=ma$ is not exact? Chapter 12-2 in Feynman Lectures Vol. 1 states:

In fact the law, $F=ma$ is not exactly true; if it were a definition we should have to say that it is always true; but it is not ...
  First, because Newton's Second Law is not exact, and second, because in order to understand physical laws, you must understand that they are all some kind of approximations. 



*

*What does he mean by approximation?  

*Also, how is $F=ma$ not a definition of force?
 A: From the context of the statement within Feynman's lectures, it is evident that what he had in mind was the idea that mathematical statements such as $F=ma$ are just an idealization of nature. In the text, he goes on to explain that for example the precise mass of a physical object is not known. He gives the example that the mass of a chair is just an approximation since we never know the exact number of atoms it contains. He then explains that mathematical definitions do not suffice to describe nature exactly, and that it is hard to define "axioms" and derive everything, since everything we will write down is just an approximation to what we measure. 
What has been written in the other answer is a good idea of what one could mean by such a statement in general, but this is not precisely what Feynman meant. 
A: F=ma is not a definition of force, because force has much more to it than just a number. What you can do is say,

"Let me define a variable x, which I will call force as mass*acceleration."

Now you come in the real world, see a body accelerating, you will say I can calculate the variable x for this body. Moreover if you give me x for this body, I can calculate its acceleration or mass. This is what mathematicians do.
Now the physicist comes and says let's observe a rope pulling a block !
The mathematician quickly calculates the value of variable x , observing the acceleration of the block, and is satisfied.

But, the physicist is interested in the "cause" of the motion of block. He asks what is similar in this block moving, and any other mass's motion?

He thinks and observes a lot of blocks moving due to different reasons and comes up with the following :-
I have seen a lot of blocks moving. Although all had different sources cause their motion, I could generalize something out of these cases. It seemed that the motion did not depend on the source , but something else that the source produced at the location of the block. Let's call this thing a dorce.

And just for the sake of calculating acceleration, let me assign a number to this thing that I have perceived as existing alongside all moving blocks that I observed.

One that produces an acceleration of $1 m/s^2$ in 1kg block gets the number 1. one that produces same acceleration in opposite direction get the number -1. One that produces an acceleration of $2 m/s^2$ is assigned the number 2 and so on..  (Note how this numbering would not exist , if he had not observed that there is a thing called dorce..in fact there would be nothing to number! )
Now the patterns he observes in the world and his numbering :-

*

*When 2 dorces which are assigned a number x and -x are made to act on a particle at the same time, it does not move! No matter what x was.

*When the dorce numbered x is made to act on 2 $1 kg$ masses tied together , half the acceleration is observed; when the force numbered x is applied to 3 $1 kg$ masses tied together , acceleration produced was $1/3$ original one, and so on. No matter what x was.

*When 2 dorces numbered x and y are made to act together on a particle, the acceleration produced was sum of that produced in case when each one acted alone.

Until and unless, you don't see what the physicist has perceived, you don't know what a dorce is. Nor will you understand the fact that what he has done is classified all the dorces of the universe and numbered each group.

Now seeing the properties 1,2,3 and his numbering he announces, To any dorce, is assigned the number d=ma where m is mass of the particle and a is the acceleration produced in the particle when the force to be numbered is applied on it and by my experiments, I propose that properties 1,2,3 hold in all cases of any dorce being applied on any body.
It is the result of the way he chose to number as well as the inherent properties of the dorce that $d=ma$ holds$*$(and is not defined) and that dorces can be summed to get net dorce.

$*$in all the cases we have observed.
Now as soon as we observe $d\neq ma$ ,i.e. that the number we assigned to the dorce doesnot equal the product of mass and acceleration produced by that dorce, we can change the number assigned to that dorce !
NOTE :- The physicist could have said, let's just define d=ma and say that this is the dorce we have perceived and let's use properties 1,2,3. Now he'd have to prove that this definition of dorce which assigns the number ma to the every dorce
that produces a acceleration in body of mass m , is consistent with rules 1,2,3.(unlike earlier case where it was proposed that all 3 properties hold) THIS CANNOT BE PROVED.  (Think why?)
Hence , the physicist will never define dorce as ma.
There have been 2 kinds of approximations in doing as in 4th blockquote :-
$1.)$The physicist doesnot measure all possible cases.(He certainly will never be able to).
$2.)$Each time he measured acceleration and mass he measured only upto certain decimal places.(He never will be able to infinite decimal places)
So all physical laws he will propose , will be some kind of approximations.
A: No, it's not exact, and it's not a definition either.  Consider that acceleration has a definition that no one will dispute.  It is the time derivative of velocity.
More than likely, he had in mind the relativistic generalization of the equation.  The more general form of the equality is:
$$ F = \frac{dp}{dt} $$
You can easily see how this results in $ma$ for the case of non-relativistic systems.  But when you use special relativity, you get a different form for momentum.
$$ p = \frac{ m_0 v }{ \sqrt{ 1 -  \left( \frac{v}{c} \right)^2 } }  $$
As you do the differentiation for the normal and relativistic forms, you need to consider that velocity is not constant over time, and its derivative is acceleration.  The rest is calculus.
Could the relativistic form of force be wrong?  Sure.  However, it is hard to imagine F=dp/dt being wrong without some contradiction popping up.  No doubt, that ties back into Noether's theorem in some way.
A: As mentioned above, newtonian-mechanics does not account for relativistic effects (nor quantum mechanical effects for that matter). This does not mean that the whole branch is wrong, it simply describes events under certain conditions ("normal" conditions, or the ones we observe in our every day life).
Regarding Feynman's comment, a derivation may come in handy.
Let $p=mv=m\frac {dx}{dt}$
Define $F=\frac {dp}{dt}$
It follows that $F=\frac {dp}{dt}=\frac {dm\frac {dx}{dt}}{dt}$
We have a product ($mv$) so we apply quotient rule to differentiate and get:
$F=\frac {dm}{dt}\cdot\frac{dx}{dt}+m\frac{d^2x}{dt^2}$
This is almost equal to the well known $F=ma$, with the exception that me have that first term $\frac {dm}{dt}\cdot\frac{dx}{dt}$. However, Newton assumed that mass is constant with respect to time, thus $\frac {dm}{dt}=0$ so it follows that $F=\frac {dm}{dt}\cdot\frac{dx}{dt}+m\frac{d^2x}{dt^2}=0 \cdot \frac {dx}{dt}+m\frac{dx^2}{dt^2}=m\frac{dx^2}{dt^2}$
Which is Newton's well known second equation.
However, as has been mentioned by others, Einstein realized that velocity is not necessarily constant over time and depends on the observer/frame of reference, thus changing the definition of momentum; as a matter of fact in some systems mass is not even constant! 
See, in relativistic terms $p = \frac{ m_0 v }{ \sqrt{ 1 -  \left( \frac{v}{c} \right)^2 } }$ and for "varying mass systems" $\frac {dm}{dt} \neq 0$
There is some heavy calculus here (I would not catalog it as hard, but rather heavy or tedious), but in the end it adds up to the following: if we consider relativist effects but start from the same definition of $F=\frac{dp}{dm}$, we conclude that $F\neq m\frac {d^2x}{dt^2}$.
Tl;dr: Because of math (and taking into consideration relativistic effects when doing math).
Edit: by the way, the relativistic force turns out to be $F=m\frac{dx^2}{dt^2}+\frac{v}{c^2}\cdot \frac{dE}{dt}$
