Since $f(x)$ is an even function defined on $\mathbb{R}$,
if $f(x)$ is an increasing function on $[0,1]$, then $f(x)$ is a decreasing function on $[-1,0]$,
Furthermore, since $f(x)$ is a function with a period of $2$ defined on $\mathbb{R}$, and $[3,4]$ is two periods away from $[-1,0]$,
the monotonicity on both intervals is consistent. Therefore, it can be concluded that $f(x)$ is a decreasing function on $[3,4]$, hence sufficiency is established.
If $f(x)$ is a decreasing function on $[3,4]$, similarly, due to the periodicity of the function, it can be concluded that $f(x)$ is a decreasing function on $[-1,0]$. Then, because the function is even, it can be concluded that $f(x)$ is an increasing function on $[0,1]$, hence necessity is established.
In summary, " $f(x)$ is an increasing function on $[0,1]$" is a necessary and sufficient condition for " $f(x)$ is a decreasing function on $[3,4]$".
Therefore, the correct answer is $\boxed{\text{D}}$.
From the problem statement, it can be deduced from the properties of the function that $f(x)$ is a decreasing function on $[-1,0]$, and then from the periodicity of the function, it can be deduced that $f(x)$ is a decreasing function on $[3,4]$, proving sufficiency. Similarly, from $f(x)$ being a decreasing function on $[3,4]$ combined with periodicity, it can be deduced that $f(x)$ is a decreasing function on $[-1,0]$, and then because the function is even, it can be deduced that $f(x)$ is an increasing function on $[0,1]$, proving necessity, which leads to the correct option.
This question examines the judgment of sufficiency and necessity. The key to solving the problem is to understand the direction of proof for sufficiency and necessity, i.e., the proof from one condition to another is sufficiency, and which direction is necessity. Beginners may get confused about the direction of the proof, leading to logical errors in expression.