1. We are given $g(x)=\int_0^x t f(t)\,dt$ and $g(x^3)=x^6+x^7.$

We need to find $\sum_{r=1}^{15} f(r^3).$

2. Let $u=x^3.$ Then the given relation becomes $g(u)=u^2+u^{7/3} \quad (u>0),$ since $x^6=(x^3)^2=u^2, \qquad x^7=(x^3)^{7/3}=u^{7/3}.$

So, $g(x)=x^2+x^{7/3}.$

3. Now differentiate using the Fundamental Theorem of Calculus. Since $g(x)=\int_0^x t f(t)\,dt,$ we have $g'(x)=x f(x).$

Also, from $g(x)=x^2+x^{7/3},$ we get $g'(x)=2x+\frac{7}{3}x^{4/3}.$

Hence, $x f(x)=2x+\frac{7}{3}x^{4/3}.$ Dividing by $x$ (for $x>0$), $f(x)=2+\frac{7}{3}x^{1/3}.$

4. Therefore, $f(r^3)=2+\frac{7}{3}(r^3)^{1/3}=2+\frac{7}{3}r.$

So, $\sum_{r=1}^{15} f(r^3)=\sum_{r=1}^{15}\left(2+\frac{7}{3}r\right).$

Split the sum: $=\sum_{r=1}^{15}2+\frac{7}{3}\sum_{r=1}^{15}r.$

Now, $\sum_{r=1}^{15}2=30,$ and $\sum_{r=1}^{15}r=\frac{15\cdot 16}{2}=120.$

Thus, $\sum_{r=1}^{15} f(r^3)=30+\frac{7}{3}\cdot 120=30+280=310.$

5. Hence the correct option is $\boxed{\text{C: }310}.$