Doubt about Taylor series: do successive derivatives on a point determine the whole function?

You're right, in general $f$ is not determined by its derivatives at one single point. Functions satisfying this condition are called analytic. But not all smooth functions are analytic, for example

$$x\mapsto\left\{\begin{array}{c}e^{-\frac{1}{x^2}}, x>0\\0, x\leq 0\end{array}\right.$$ is a smooth function and the derivatives at zero are all zero, hence the Taylor series developed at zero does not determine the function.

Furthermore the exact statement of Taylor's theorem is quite different from what you said. It is as follows:

If $f\in C^{k+1}(\mathbb{R})$, then $$f(x)=\sum_{n=0}^k f^{(n)}(a)(x-a)^n\frac{1}{n!} + f^{(k+1)}(\xi)\frac{1}{(k+1)!}(x-a)^{k+1}$$

If you now take $k\rightarrow\infty$ it is in general not clear, that this error term converges to zero.

Functions which are the sum of their Taylor series within the interval (or disk for functions of a complex variable) of convergence are known as analytic functions. Many basic elementary functions are analytic: $\;\exp, \sin,\cos,\sinh,\cosh $ and of course polynomials are analytic on $\mathbf R$ (or $\mathbf C$).

It is not true that, in general, an infinitely differentiable function of a real variable is analytic on the interval of convergence of its Taylor series, as @humanStampedist's example shows.

However, for a function of a complex variable, simply being differentiable suffices to ensure the function is analytic (one usually says holomorphic in this case). This is due to the very strong constraints of the Cauchy-Riemann equations.

HumanStampedist has adequately answered the question. I'd like to mention that, just as there are continuous functions that are nowhere differentiable, such as the Weierstrass function, there are smooth functions (all $n$th derivatives exist at every point) that are nowhere analytic, i.e. at no point does the Taylor series converge to the original function. An example is the Fabius function.