Poisson process can be interpreted as a continuous case of Bernoulli process.
Taking your example, consider that the buyer is consuming the good in batches in fixed intervals of time with probability $p$. So the consumption is allowed only after fixed intervals, and the r.v. $X_t \sim Bern(p)$, where $X_t=1$ indicates that the buyer consumes the good at time $t$. Further, how much she consumes in $n$ periods is distributed as $Bin(n,p)$. The expected level of consumptions in $n$ periods as we know is $np$.
Now consider that the buyer can consume at any instant and just not in intervals. Unlike in previous case where consumption could happen only at $t=1,2,...n$, now it can take place at any $t\in \mathbb{R^+}$. Naturally the probability that consumption happens at some given $t$ is $0$. So comes the rate $\lambda$, i.e., we say that the buyer on average consumes $\lambda$ amount in a given length $n$ of time. Now the total consumption between $t=0$ to $t=n$ becomes $Pois(\lambda)$.
So, $Pois(\lambda)$ is a limiting case of $Bin(n,p)$ as time interval $\to 0$.
Coming to your second question:
The buyer gets value $v$ when he uses the good at a given time, and the value gets depreciated after one time interval by $\delta$. The total value, as you mention, is: $v(1+\delta+\delta^2+..)$
However, if we assume the value is depreciating continuously at rate $r$ we can say:
\begin{align}
rv(t)&=\lim_{\Delta t \to 0} \frac{v(t+\Delta t) - v(t)}{\Delta t} \\
&=\frac{dv(t)}{dt} \\
\end{align}
From this you will get that the value from having the good over a time interval $[0,d]$ be $\int_0^d v e^{-rt} dt$
Lastly, to answer the last part: if you want to go to Bernoulli process from the described poisson process consider that arrivals can happen only in discrete time intervals such that on average in $d$ periods, $\lambda_d$ arrivals take place. So, this translates to the bernoulli process such that arrivals happens at a given time $t$ with probability $p=\lambda_d/d$.