Let $k$ be a field, $a\in k^\times$, and let’s assume that $k$ contains all $n$-roots of unity for $n \in \mathbb{N}$ (let $\mu_n$ be the group of $n$-roots of unity in $k$, then this means that $|\mu_n|=n$). Let $K$ be the extension of $k$ generated by an element $\alpha$ such that $\alpha^n=a$. Then $K$ is the decomposition field of the polynomial $X^n-a$ over $k$, and the extension $K/k$ is Galois.
According to my lecture notes, there’s an injective group homomorphism $\operatorname{Gal}(K/k) \rightarrow \mu_n$ that sends every $\sigma \in \operatorname{Gal}(K/k)$ to a root of unity $\xi^\sigma$. We then get an isomorphism $\operatorname{Gal}(K/k) \cong \mu_m$, where $m|n$.
I understand all of this so far, however, I find it difficult to imagine intuitively how could $m$ be strictly inferior to $n$. Maybe an example could help, if someone could provide one.
The next section of my lecture notes I find harder to understand. It says:
The element $\prod_\sigma \sigma(\alpha)=\alpha^m \prod_{\sigma}\xi_\sigma$ is fixed by $\operatorname{Gal}(K/k)$, then pertains to $k$. Hence $a=(\alpha^m)^{n/m}\in (k^\times)^{n/m}$, and we obtain that $m$ is the order of $a$ in $(k^\times)^{n/m}$, and in particular that the polynomial $X^n-a$ is irreducible in $k[X]$ iff the order of $a$ in $(k^\times)^{n/m}$ is $n$.
I don’t understand how to get to the conclusion in the last sentence (what I wrote in bold). Any help would be appreciated.
Let's start with your first question. Consider the extension $K/k$ where $\mu_n := \langle \zeta_n \rangle \subset K = k(\alpha)$. The extension $K/k$ is indeed Galois, as $K$ is a splitting field of $f:= x^n-a \in k[x]$ (assume that char$(k)\not \mid n$, otherwise the extension might not be separable; for example $x^{2p}-3 = (x^2-3)^p$ in char$(k) = p>0$).
It is important to notice that $f = \prod_{k=1}^n (x-\zeta_n^k \alpha)$. Consider the map: $\psi: \text{Gal}(K/k) \to \mu_n$ defined by sending $\sigma \mapsto \frac{\sigma(\alpha)}{\alpha}$, which is indeed well-defined. It is an injective homomorphism of groups. Thus by the isomorphism theorem it follows that $\text{Gal}(K/k) \cong \psi[\text{Gal}(K/k)] \subset \mu_n$.
Remember that $\langle \zeta_n \rangle$ has for every divisor $l$ of $n$ a unique subgroup of index $l$, and it is of the form $\langle \zeta_n^l \rangle$. Thus $\psi[\text{Gal}(K/k)] = \langle \zeta^k_n \rangle$ for some divisor $k$ of $n$ (let's say $n = km$). Notice that $\zeta^k_n$ is a primitive $m$-th rooth of unity, therefore $\psi[\text{Gal}(K/k)] = \mu_m$. (I have written this so you can see that $m$ is not always equal to $n$; notice that when $n$ is prime it will be equal to $m$, if you have a non-trivial extension $K/k$.)
Notice that when $f$ is irreducible over $k$, then $\# \text{Gal}(K/k) = n$. So to find an example where $m<n$ you should consider reducible polynomials over $k$. For example, take $k = \mathbf{Q}$ and $f:= x^4-2^4 \in \mathbf{Q}[x]$, then you can see that $\text{Gal}(K/k) \cong C_2$.
For the second question, I don't understand what the order of $a$ in $(k^{\times})^{n/m}$ means. But I think what the last sentence actually means is that $a$ is 'really' an $n$-th power of $\alpha$, that is the biggest divisor $m$ of $n$ such that $\alpha^m = a$ is equal to $n$. If you rephrase it this way, the last sentence makes more sense to me.
However, using your terminology, if $f$ is irreducible then it follows that $m = n$ (because $\text{Gal}(K/k) \cong \mu_m$, where $\# \text{Gal}(K/k) = n$). And, if the order of $a$ in $(k^{\times})^{n/m}$ is equal to $n$, then it follows by the sentence before the sentence in bold that $\# \text{Gal}(K/k) = n$. Therefore, $n\mid \# \text{Gal}(K/k)$ and this can only be the case when $f$ is irreducible.