Poisson basic process formula Let, $N_t$ be a Poisson process  and let $X_t$ solve the SDE $d{X_t}=a_t dt +J_t dN_t$.  Then,  Ito´s fórmula is:
$$df(t,X_t)=(\frac{\partial f}{\partial t} +  \frac{\partial f}{\partial x}a_t)dt + (f(t, X_t+J_t)-f(t,X_t))dN_t$$
I don´t know how to demonstrate it.
Note: I understand that when there is a jump $N_t$ jumps 1 unit, so $X_t$ jumps $J_t$ and hence $f$ jumps $f(t, X_t+J_t)-f(t,X_t)$ when there is a ump at time t. 
 A: As you mentioned in your note, the outcome of the jump process part of the sde is a step functions (of times) that jumps up a distance $J_{t_i}$ at a sequence of random times $t_i$ which are an outcome of the Poisson process $N_t$.  
Let's just consider the SDE: $dX_t = J_tdN_t$.
Then given one of the outcomes I mentioned above, $f(X_t)$ is simply $f$ evaluated at where $X_t$ is at that moment of time.  At an exact jump moment $t_i$, this will be $f(X_{t_i} + J_{t_i})$.  Since before the jump we were at the value $f(X_{t_i})$ we need to add the difference to our current location to get our next location.
$$f(X_t) = f(X_{t_i}) + (f(X_{t_i} + J_{t_i}) - f(X_{t_i}))$$
for all $t > {t_i}$ but before the next jump $t_{i+1}$.
If right this in terms of all the jumps that have occurred we have:
$$f(X_t) = f(X_0) + \sum_{k=0}^i (f(X_{t_i} + J_{t_i}) - f(X_{t_i}))$$
We can rewrite each summand as time integral by integral a dirac mass at time $t_i$:
$$f(X_t) = f(X_0) + \sum_{k=0}^i \int_0^t (f(X_s + J_s) - f(X_s)) \delta_{t_i}(s) ds$$
and this equals
$$f(X_0) + \int_0^t (f(X_s + J_s) - f(X_s)) \sum_{k=0}^i \delta_{t_i}(s) ds$$
Now if we defined $dN_t$ to be a measure with dirac masses at time $t_i$ we get:
$$f(X_t) = f(X_0) + \int_0^t (f(X_s + J_s) - f(X_s)) dN_s $$
In other words, this is just a very elaborate way of setting up a telescoping sum to form step functions.
