1. Given data

We have $F(x)=x f(x), \qquad f(2)=1.$ So, $F(2)=2f(2)=2.$

Also given: $\int_0^2 xF'(x)\,dx=6$ and $\int_0^2 x^2F''(x)\,dx=40.$

We need to find $F'(2)+\int_0^2 F(x)\,dx.$

2. Use integration by parts on $\int_0^2 xF'(x)\,dx$

Let us evaluate $I_1=\int_0^2 xF'(x)\,dx.$

Using integration by parts: $\int xF'(x)\,dx = xF(x)-\int F(x)\,dx.$ Hence, $\int_0^2 xF'(x)\,dx = \left[xF(x)\right]_0^2-\int_0^2 F(x)\,dx.$

Since $F(2)=2$ and $xF(x)$ at $x=0$ is $0$, $6 = 2\cdot F(2)-\int_0^2 F(x)\,dx = 2\cdot 2-\int_0^2 F(x)\,dx.$ So, $6=4-\int_0^2 F(x)\,dx,$ which gives $\int_0^2 F(x)\,dx = -2.$

3. Use integration by parts on $\int_0^2 x^2F''(x)\,dx$

Let $I_2=\int_0^2 x^2F''(x)\,dx=40.$

Integrating by parts: $\int x^2F''(x)\,dx = x^2F'(x)-\int 2xF'(x)\,dx.$ Therefore, $\int_0^2 x^2F''(x)\,dx = \left[x^2F'(x)\right]_0^2 -2\int_0^2 xF'(x)\,dx.$

Now, $\left[x^2F'(x)\right]_0^2 = 4F'(2),$ and $\int_0^2 xF'(x)\,dx=6.$ Thus, $40=4F'(2)-2(6)=4F'(2)-12.$ So, $4F'(2)=52 \implies F'(2)=13.$

4. Compute the required expression

We need $F'(2)+\int_0^2 F(x)\,dx = 13+(-2)=11.$

5. Final answer

$\boxed{11}$ So the correct option is B.