Why do additional Taylor terms lead to an improved approximation of a function? I hope my question becomes clear:
I understand what a Taylor polynomial does. It approximates an analytical function in the point $x=a$ in a way that the n'th order Taylor polynomial matches the function up to its n'th derivative in the point $x=a$.
Something I could never answer myself though is:
Why does each additional Taylor term improve the approximation of the function in the vicinity of the point $x=a$? Can't it be that a Taylor polynomial provides a worse approximation in the vicinity of $x=a$ when taking more Taylor terms (i.e. choosing a later truncation for $n$)?
If this is the case:
Is it also safe to say that truncating a Taylor series at larger $n$ also improves the approximation of $f(x)$ for points far away from $x=a$?
And a last question:
Why is an analytic function $f(x)$ almost always equal to the Taylor series of the form
$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}}{n!} (x-a)!$
Are there function where this is not the case? And is there a clear cut proof, why an infinite Taylor series equals an analytic function in general?
 A: 
Why does each additional Taylor term improve the approximation of the function in the vicinity of the point $x=a$?

It does not always do that.

Can't it be that a Taylor polynomial provides a worse approximation in the vicinity of $x=a$ when taking more Taylor terms (i.e. choosing a later truncation for $n$)? 

It could be the case that taking some additional terms provides a worse approximation.
The idea is merely that the series eventually converges to the correct value.
So you will get a better approximation at whatever point you are looking at (within the radius of convergence) if you take enough additional terms.

Is it also safe to say that truncating a Taylor series at larger $n$ also improves the approximation of $f(x)$ for points far away from $x=a$?

Not always. A particular increase in $n$ might improve the approximation at a closer point but make it worse farther away.
For example, consider the Taylor series for $\cos(x)$ about $x = 0.$
Consider the Taylor polynomials
\begin{align}
p_0(x) &= 1,\\
p_2(x) &= 1 - \frac12 x^2,\\
p_4(x) &= 1 - \frac12 x^2 + \frac1{24}x^4,\\
p_6(x) &= 1 - \frac12 x^2 + \frac1{24}x^4 - \frac1{720}x^6.\\
\end{align}
Now evaluate these at $x= 1$. We get
\begin{align}
p_0(1) &= 1,\\
p_2(1) &= 0.5,\\
p_4(1) &\approx 0.54167,\\
p_6(1) &\approx 0.54028,\\
\end{align}
each of which gets progressively closer to $\cos(1),$ which is
approximately $0.54030.$
But at $x= 5$ we get
\begin{align}
p_0(5) &= 1,\\
p_2(5) &= -11.5,\\
p_4(5) &\approx 14.54,\\
p_6(5) &\approx -7.16,\\
\end{align}
whereas $\cos(5)$ is approximately $0.28.$
So we see that from $p_0$ to $p_4$ the approximation just keeps getting worse,
and it does not even begin to improve until $p_6.$
Continuing with higher-order polynomials, $p_8(5) \approx 2.52$ and
$p_{10}(5) \approx -0.16$.
The absolute error of $p_{10}$ is a little less than $0.45,$ which is the first time we get an error less than the error of $p_0,$ which is about $0.72.$
Now, whether $x = 5$ is really in the "vicinity" of $x = 0$ is something you might argue, but considering that the radius of convergence of the Taylor series about $x = 0$ is infinite, $x = 5$ is not really that far away.
And we could always do a similar analysis for a function such as 
$\frac1{10000}\cos(10000x),$ for which the behavior of the Taylor polynomials at $x = 0.0005$ is analogous to the behavior we examined at $x = 5$ above.

Why is an analytic function $f(x)$ almost always equal to the Taylor series of the form
  $$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}}{n!} (x-a)^n ?$$
  Are there function where this is not the case?

I would not say "almost always." There are some "nice" functions such as polynomials or sinusoidal functions whose radius of convergence is infinite;
in general, however, we have a finite radius of convergence, which means that the Taylor series that you find around a particular point is wrong
(in fact does not even converge) on a far greater part of the number line than the part on which it is correct.
On the other hand, if by "almost always" you just mean that the Taylor series is almost always correct within some neighborhood of the point we take it about,
you can delete the word "almost." By definition, if a function $f$ is a real analytic function then at every real number $x_0$ the Taylor series of $f$ around $x=x_0$ is correct on some neighborhood of $x_0.$

And is there a clear cut proof, why an infinite Taylor series equals an analytic function in general?

The proof is by definition.
A more interesting question is whether an infinitely differentiable function always has a Taylor series at every point that is accurate on an interval about that point.
The answer is no.
An example that is often cited is
$$
f(x) = \begin{cases} e^{-1/x^2} & x\neq 0, \\ 0 & x = 0, \end{cases}
$$
whose Taylor series around $x = 0$ is simply zero, which is the correct value of the function only at the single point $x = 0$ itself.
See this answer for further discussion.
