I would like to calculate the Doob-Meyer decomposition of $(N_l-\lambda l)^2$ ($l\geq0$), and $N_l$ is a Poisson process with parameter $\lambda>0$.
I know that $(N_l-\lambda l)^2$ is non-negative, right continues submartingale, so there exists the D-M decomposition, but how can I get it?
Let's define $M_t := N_t - \lambda t$, which is a martingale as I think you implicitely noted. Also (and this is quite crucial), $M$ has independent stationary increments.
The answer to your question is that the increasing previsible process in the DM decomposition of $M^2$ is $A_t = \lambda t$, which amounts to saying that $(M_t^2 - \lambda t)_t$ is a martingale.
To show this, just note that $$ E_t (M_{t+h}^2 - M_t^2) = E_t (M_{t+h} - M_t)^2 = E (M_{t+h} - M_t)^2 = E M_{h}^2 $$ where the first equality is true for every square integrable martingale, the second follows from the independence of the increments and the third from their stationarity.
We can easily compute $E M_{h}^2$: $$ E M_{h}^2 = \text{var} (M_{h}) = \text{var} (N_{h} - \lambda h) = \text{var} (N_{h}) = \lambda h $$
Putting the pieces together we showed that $$ E_t (M_{t+h}^2 - M_t^2) = \lambda (t+h) - \lambda t $$ which is just the same thing as saying that $(M_t^2 - \lambda t)_t$ is a martingale.
As an aside, the increasing previsible process in the DM decomposition of the square of an $L^2$ martingle is denoted by $\langle M \rangle$ and is referred to as the angle bracket process of $M$. So your question could have been just: "What is the angle bracket of a compensatad Poisson Process?".