By the mean value theorem $$|f(t_i)-f(t_{i-1})|=f'(c)(t_{j}-t_{j-1})$$
and Let the total variation be $$sup\sum_{j=1}^{n-1}|f_{t}-f_{t-1}|$$
Now.Taking norm of the partition $$\left\|∏\right\| = max[(t_{j}-t_{j-1}):i=1,2....(n)]$$ By definition (it's intuitive) that
$$\left\|∏\right\|\ge(t_{j}-t_{j-1})$$ Therefore, replacing and / or combining with the mean value theorem.
$$\sum_{j=1}^{n-1}f'(c)\left\|∏\right\|\ge\sum_{j=1}^{n-1}f'(c)(t_{j}-t_{j-1})$$
Now. Taking the left part and taking limit on the maximum partition tending to zero. $$lim_{∏\rightarrow0}\left\|∏\right\|\sum_{j=1}^{n-1}f'(c) = 0$$
I know that the answer must not be zero, but a definite integral over the considered interval. I am intrigued to know why this test is correct for quadratic variation, but not for total variation.
This could be a comment but is too long.
There seems to be at least four errors in your analysis. First, your use of the mean value theorem should be without the modulus signs or you should include modulus signs on the derivative as well. Second, when deriving the inequality $$\sum_{j\geqslant 1} f'(c_j)\lVert \Pi \rVert \geqslant \sum_{j\geqslant 1} f'(c_j)(t_j - t_{j-1})$$ you seem to be assuming each $f'(c_j)$ is positive. These error would be avoided if you used $\lvert f'(c_j)\rvert$ throughout. Third, in your notation the value $c$ depends on the interval so you should write it in a way that makes clear each occurrence of $c$ in the sum is different, e.g. by writing $c_j$.
But fourth, and most significantly, as observed in comments, when $\lVert \Pi \rVert \to 0$, $n \to \infty$ and so you have a longer and longer sum of (possibly) smaller and smaller terms. However that does not guarantee convergence to zero. You need only consider $f(x) = x$ to obtain a counter example, where the right hand side of your inequality is always exactly equal the length of the interval and the left side always something greater, so neither side has limit zero.