We are given $f(x)+3f\left(\frac{24}{x}\right)=4x, \qquad x\neq 0.$ We need to find: $f(3)+f(8).$

Substitute $x=3$: $f(3)+3f\left(\frac{24}{3}\right)=4\cdot 3$ $f(3)+3f(8)=12. \qquad (1)$

Substitute $x=8$: $f(8)+3f\left(\frac{24}{8}\right)=4\cdot 8$ $f(8)+3f(3)=32. \qquad (2)$

Let $a=f(3), \qquad b=f(8).$ Then equations (1) and (2) become: $a+3b=12$ $3a+b=32$

Solve the system. From the first equation, $a=12-3b.$ Substitute into the second: $3(12-3b)+b=32$ $36-9b+b=32$ $36-8b=32$ $-8b=-4$ $b=\frac{1}{2}.$ Then $a=12-3\left(\frac{1}{2}\right)=12-\frac{3}{2}=\frac{21}{2}.$

Therefore, $f(3)+f(8)=a+b=\frac{21}{2}+\frac{1}{2}=\frac{22}{2}=11.$

Hence the correct option is: $\boxed{\text{B: }11}$