$$P(N_1=2, N_4=6) = P(N_1=2, N_4-N_1=4) = P(N_1=2, N_3=4) = P(N_1=2) \cdot P(N_3=4)$$
What I understand is:
$P(N_1=2, N_4=6)$ means "the probability of count of 2 items arriving at step-1, AND 6 items arriving at step-4.
$P(N_1=2, N_4-N_1=4)$ means that "the probability of count of 2 items arriving at step-1 AND the 4 items as the difference of steps 4 and 1.
which is same as $P(N_1=2, N_3=4)$ i.e. "the probability of count of 2 items at step-1 AND 4 items at step-3".
I understand (2), as (1) is represented as a "difference" term in (2).
I also understand that the increments of a Poisson Process are independent. Therefore, $$P(N_1 = 2, N_4-N_1=4)= P(N_1=2)P(N_4-N_1=4).$$
But, how are $P(N_1=2, N_4-N_1=4)$ and $P(N_1=2, N_3=4)$ equivalent?
Note that actually $P(N_1=2, N_4-N_1=4) \color{red}{\ne}P(N_1=2, N_3=4)$. In fact, we have $$P(N_1=2, N_3=4) = P(N_1=2)P(N_2=2),$$ while $$ P(N_1 = 2, N_4-N_1=4) =P(N_1=2)P(N_3=4).$$
These are equal if and only if $P(N_2=2)=P(N_3=4)$, i.e. the rate $\lambda$ satisfies $$e^{-2\lambda}\frac{4\lambda^2}{2!}=e^{-3\lambda}\frac{81\lambda^4}{4!}.$$
This is the case if and only if $e^{\lambda} = \frac{81}{48}\lambda^2$. As this plot shows, there is no $\lambda >0$ that satisfies this.
A correct simplification is as follows. Note that the increments of a Poisson process are independent. So $N_1$ and $N_4-N_1$ are independent. Therefore, $$P(N_1 = 2, N_4-N_1=4)= P(N_1=2)P(N_4-N_1=4).$$ Also, by stationary increments of a Poisson process, $N_4-N_1$ has the same distribution as $N_3$. (In general, $N_t-N_s$ has the same distribution as $N_{t-s}$ for any $0\le s \le t$). Hence $$P(N_4-N_1=4)= P(N_3=4).$$
Combine these facts together to get $$P(N_1 = 2, N_4-N_1=4) =P(N_1=2)P(N_3=4).$$