The part of the Frequentist approach that clashes with the likelihood principle is the theory of statistical testing (and p-value computation). It is usually highlighted by the following example.
Suppose two Frequentist want to study a biased coin, which turns 'heads' with unknown propability $p$. They suspect that it is biased towards 'tail', so they postulate the same null hypothesis $p = 1/2$ and the same alternative hypothesis $p < 1/2$.
The first statistician flips the coin until 'heads' turns up, which happens to be 6 times. The second decides to flip the coin 6 times, and obtains only one 'heads' in the last throw.
According to the model of the first statistician, the p-value is computed as follows:
$$ p(1-p)^5 + p(1-p)^6 + ... = p(1-p)^5 \frac{1}{1-p} = p(1-p)^4. $$
According to the model of the second statistician, the p-value is computed as follows:
$$ {6 \choose 1} p(1-p)^5 + {6 \choose 0} (1-p)^6 = (5p + 1)(1-p)^5. $$
Replacing $p$ by $1/2$, the first finds a p-value equal to $1/2^5 = 0.03125$, the second finds a p-value equal to $7/2 \times 1/2^5 = 0.109375$.
So, they get different results because they did different things, right? But according to the likelihood principle, they should come to the same conclusion. Briefly, the likelihood principle states that likelihood is all that matters for inference. So the clash here comes from the fact that both observations have the same likelihood, proportional to $p(1-p)^5$ (likelihood is determined up to a proportionality constant).
As far as I know, the answer to your second question is more of a debated opinion. I personally try to avoid performing tests and computing p-values for the reason above, and for others explained in this blog post.
EDIT: Now that I think about it, estimations of $p$ by confidence intervals would also differ. Actually if the models are different, the CI differ by construction.