To analyze whether "$0 \lt \lambda\ \ \lt 2$" is a sufficient and/or necessary condition for the sequence $\{a_{n}\}$ to be an increasing sequence, we start by examining the difference between consecutive terms of the sequence:

$$\begin{align*} a_{n+1}-a_{n} &= \left(n+1+\frac{\lambda}{n+1}\right) - \left(n+\frac{\lambda}{n}\right) \\ &= 1 + \frac{\lambda}{n+1} - \frac{\lambda}{n} \\ &= 1 - \frac{\lambda}{n(n+1)} \\ &= \frac{n^2 + n - \lambda}{n(n+1)}. \end{align*}$$

Given that the quadratic function $f(n) = n^2 + n - \lambda$ has its axis of symmetry at $n = -\frac{1}{2}$, it is monotonically increasing for $n \geq 1$. Therefore, the minimum value of $f(n)$ within the domain of interest occurs at $n=1$, which is $f(1) = 2 - \lambda$. Since $0 \lt \lambda \lt 2$, we have:

$$f(n) \geq f(1) = 2 - \lambda \gt 0.$$

This implies that $a_{n+1} - a_{n} \gt 0$ for all $n$, indicating that the sequence $\{a_{n}\}$ is indeed an increasing sequence when $0 \lt \lambda \lt 2$. Thus, "$0 \lt \lambda \lt 2$" is a sufficient condition for the sequence to be increasing.

To check if it's a necessary condition, consider $\lambda = 0$. In this case, $a_{n} = n$, which is clearly an increasing sequence. This shows that the condition "$0 \lt \lambda \lt 2$" is not necessary for the sequence to be increasing, as the sequence can still be increasing outside this range of $\lambda$.

Therefore, "$0 \lt \lambda \lt 2$" is a sufficient but not necessary condition for the sequence $\{a_{n}\}$ to be an increasing sequence.

Hence, the correct answer is $\boxed{\text{A}}$: A sufficient but not necessary condition.