From the given equation, we have $x^{4}+a_{1}x^{3}+a_{2}x^{2}+a_{3}x+a_{4}=(x+1)^{4}+b_{1}(x+1)^{3}+b_{2}(x+1)^{2}+b_{3}(x+1)+b_{4}$.

Substitute $a_{1}=2$, $a_{2}=0$, $a_{3}=1$, and $a_{4}=6$ into the equation, we get $x^{4}+2x^{3}+x+6=(x+1)^{4}+b_{1}(x+1)^{3}+b_{2}(x+1)^{2}+b_{3}(x+1)+b_{4}$.

Now, let's find the value of $f(2,0,1,6)$, which is $b_{1}-b_{2}+b_{3}-b_{4}$. To do this, we need to find the values of $b_{1}$, $b_{2}$, $b_{3}$, and $b_{4}$.

Let's take $x=-2$ in the equation $x^{4}+2x^{3}+x+6=(x+1)^{4}+b_{1}(x+1)^{3}+b_{2}(x+1)^{2}+b_{3}(x+1)+b_{4}$. This simplifies to $b_{1}-b_{2}+b_{3}-b_{4}=-3$.

Therefore, the value of $f(2,0,1,6)$ is $\boxed{-3}$.