First of all, there is not *the* algebraic/separable closure. Choices have to be made. However, if an algebraic closure $k^{\mathrm{alg}}$ of $k$ is fixed, inside it there is a unique separable closure $k^{\mathrm{sep}}$ of $k$, namely the subfield consisting of the separable elements over $k$.

Ignoring the failure of uniqueness, you can regard $k^{\mathrm{alg}}$ as the biggest algebraic extension of $k$, whereas $k^{\mathrm{sep}}$ is the biggest galois extension of $k$. The latter is because $k^{\mathrm{sep}}$ is easily seen to be normal. In particular, you can apply Galois theory and relate the group theory of the absolute Galois group $\mathrm{Gal}(k^{\mathrm{sep}}/k)$ with the field theory of Galois extensions of $k$. The algebraic closure is too big to make Galois theory work.

Obviously $k$ is perfect if and only if $k^{\mathrm{alg}} = k^{\mathrm{sep}}$. Finite fields and fields of characteristic $0$ (in particular number fields) are perfect. But what is the difference in the other cases? Let $p = \mathrm{char}(k) > 0$. Then $k^{\mathrm{alg}} / k^{\mathrm{sep}}$ is purely inseparable, i.e. for every $a \in k^{\mathrm{alg}}$ there is some $n \geq 1$ such that $a^{p^n} \in k^{\mathrm{sep}}$. In other words, this field extension is given by adjoining all $p^n$-th roots. A consequence of this is that the restriction map $\mathrm{Aut}_k(\overline{k}) \to \mathrm{Gal}(k^{\mathrm{sep}}/k)$ is an isomorphism.

Actually one can show that the canonical map $k^{\mathrm{sep}} \otimes_k k^{\mathrm{perf}} \to k^{\mathrm{alg}}$ is an isomorphism, where $k^{\mathrm{perf}}=\cup_{n \geq 0} k^{1/p^n}$ is the perfect hull of $k$.