$\mathrm{d}f(x,t)$ this way $d\big(\,f(t,x)\big)=\frac{\partial f}{\partial t} \,dt+\frac{\partial f}{\partial x}\,dx$? 
*

*If $dX_t=a_t \,dt$ the next procedure is correct?


$$\mathrm{d}\big(\,f(t,x)\big)=\frac{\partial f}{\partial t} dt+\frac{\partial f}{\partial x}dx=\frac{\partial f}{\partial t} dt+\frac{\partial f}{\partial x}a_t\,dt$$


*And, if it is correct, why it can´t be applied when $X_t=a_t\, dt + b_t
\, dN_t$ or $X_t=a_t\, dt + b_t\, dW_t$?


($N_t$ is a Poisson process and $W_t$ is the Brownian motion)
Even if you only know answer (1), I would appreciate.
 A: You have a term corresponding to changes in $f$ caused by changes in $t$ and a term corresponding to changes in $f$ caused by changes in $x$, where the changes in $x$ are themselves caused by changes in $t$. When $x$ is differentiable in time, you get what you wrote by using the classical chain rule. In stochastic calculus $x$ is not differentiable in time, so the classical chain rule does not apply. Ito's formula is the generalization of the classical chain rule for stochastic calculus. 
There are a number of ways to argue that the classical chain rule shouldn't work in stochastic calculus. One is to take as given that an Ito integral should be a martingale (which itself can be justified in several ways). Then if the chain rule worked, you would have $\int_0^t W_s dW_s = \frac{W_t^2}{2}$, which is not a martingale (its expectation grows in time). Instead you have an additional "correction term" which simply takes away the drift: $\int_0^t W_s dW_s = \frac{W_t^2-t}{2}$.
A heuristic explanation which is nice for remembering formulas and doing calculations is that in regular calculus, $(dt)^2=0$ (this can be rigorously explained with Riemann sums) but $(dW)^2$ or $(dN)^2$ are not zero (this also can be explained with Riemann sums).
I'll do the calculation for the first one:
$$\int_0^t (ds)^2 := \lim_{n \to \infty} \sum_{k=0}^{n-1} (t/n)^2 = \lim_{n \to \infty} t^2/n = 0.$$
I'll write the second one, and the result, and leave you to prove it:
$$\int_0^t (dW_s)^2 := \lim_{n \to \infty} \sum_{k=0}^{n-1} \left ( W_{(k+1)t/n}-W_{kt/n} \right )^2 = t.$$
In this sense "$(dW_t)^2=dt$".
A: For (1), I'm guessing you meant $X_t=a_t$ where $a_t$ is differentiable wrt $t$, in which case you are correct.
You can't apply this to (2) because the Poisson process doesn't have zero quadratic variation (which is a property of $a_t$ in (1)); instead you need to use Ito's formula (with jumps), which does not reduce to the formula you obtained in (1).
Ito's formula without jumps is $$\begin{align}df&=\frac{\partial f}{\partial t} dt+\frac{\partial f}{\partial x}dx+\frac{1}{2}\frac{\partial^2 f}{\partial x^2}(dx)^2\end{align}$$
which you could simplify further by plugging in what you have for $dX_t$. Notice that for (1), $(dX_t)^2$ is of order $(dt)^2$ (and can therefore be taken to be zero; look up "Ito's multiplication rule" for more details), whereas this is not true for (2).
