Showing that the language $L = \{\langle M, w \rangle\ |\ M$ moves left at least three times while computing $w \}$ is decidable or undecidable
I strongly believe it is decidable, but I found it quite convoluted. I'll give a sketch of proof.

Lemma: Every Turing Machine that doesn't terminate either moves left at some point or is very silly.

Let $M_\leftarrow$ be a TM that never moves left. If $M_\leftarrow$ never moves left, we don't care about what it writes on the tape, because it can only ever read the last symbol it wrote. In particular, we may see $M_\leftarrow$ as a finite transition system which set of states are the pairs $(q \in Q, \sigma \in \Sigma)$ where $q$ is the current state of $M_\leftarrow$ and $\sigma$ is either the last symbol written by $M_\leftarrow$ or the blank symbol if $M_\leftarrow$ just moved to the right. Therefore, checking if the graph of the transition system is cyclic decides whether or not $M_\leftarrow$ terminates, qed.

Theorem: $L$ is decidable.

Let $H$ be the problem of deciding whether or not a machine that never moves left terminates. By the lemma above, there exists a machine $M_H$ that takes as input another machine $M$, some input $x$ and:


if $M$ never moves left on input $x$, returns $\textsf{YES}$ if $M$ terminates and $\textsf{NO}$ otherwise,
if $M$ on input $x$ moves left at some point, returns $\textsf{YES}$.


We will show that we can use $M_H$ to construct a family of machines $M_k$ that decides whether or not a machine moves left at least $k$ times. Let $\langle M, x \rangle$ be the input. 

First of all, we have $M_k$ construct a universal machine $U$, and use $U$ to compute $M_H$ on input $M$. If the result is $\textsf{NO}$, it means that $M$ never moves left on input $x$: we accept $\langle M, x \rangle$. If the result is $\textsf{YES}$, it means that either $M$ terminates on input $x$ or moves left at some point. 

We use $U$ to simulate $M$ on input $x$. After finitely many steps, either $M$ terminates or moves left. In the first case simply accept. If a left move happens, we save the current state of the tape, and return the result of $M_{k-1}$ with the modified tape as input, which is by inductive hypothesis computable, qed.