This is a very nice question (and indeed, I remember asking myself this question when I was a graduate student).

The answer in general is that no, you do not get any extra strength from having these small covering sets for strong compactness. Indeed, I claim that one can have such small covering sets for every regular $\mu$ above $\kappa^+$, while $\kappa$ is the least measurable cardinal and exhibits no nontrivial degree of supercompactness.

To see this, consider the usual term-forcing proof of Magidor's theorem showing that the least measurable cardinal can be strongly compact. (This is different from his original argument using iterated Prikry forcing, and he never published it, but it has appeared in numerous articles by Apter, myself and others. See for example, A. W. Apter and J. D. Hamkins, “Exactly controlling the non-supercompact strongly compact cardinals,” J. symbolic logic, vol. 68, iss. 2, pp. 669-688, 2003.) Let me sketch it, but you may want to look at the fuller accounts that are available. Start with $\kappa$ supercompact in $V$, plus GCH, with no measuralbe cardinals above $\kappa$. Let $\mathbb{P}$ be the Easton-support $\kappa$-iteration that at each measurable cardinal $\gamma<\kappa$ forces to add a stationary non-reflecting set $S_\gamma\subset\gamma\cap\text{Inacc}$. Let $G\subset\mathbb{P}$ be $V$-generic and consider $V[G]$. All the measurable cardinals $\gamma<\kappa$ have been killed, and it follows by a theorem of mine that no new measurable cardinals are created. To see that $\kappa$ is strongly compact, fix any large regular $\mu$ above $\kappa$, and let $j_0:V\to M_0$ be a $\mu$-supercompactness ultrapower. Let $h:M_0\to M$ be a Mitchell-minimal normal ultrapower, so that $\kappa$ is not measurable in $M$. Let $j=h\circ j_0:V\to M$ be the composition. The next stage of forcing in $j_0(\mathbb{P})$ above $\kappa$ is beyond $\mu$. The term-forcing for $j_0(\mathbb{P})_{\kappa+1,j_0(\kappa)}$ over the stage $\kappa$ forcing is $\leq\mu$-closed, and so we can diagonalize to produce an $M[G]$-generic $G_{term}$ for that. Now, $j(\mathbb{P})=\mathbb{P}*\mathbb{P}_{\kappa,h(\kappa)}*\mathbb{P}_{h(\kappa),j(\kappa)}$, and there is no forcing at stage $\kappa$ since it isn't measurable in $M$. One can use the usual diagonalization methods to produce an $M[G]$-generic filter $G_{h(\kappa)+1}$ for the forcing up and including stage $h(\kappa)+1$. Next, $h''G_{term}$ is $M$-generic for the term forcing of the tail forcing over $M[G_{h(\kappa)+1}]$, and so one can construct the rest of the generic filter and thereby get $M$-generic $j(G)\subset j(\mathbb{P})$ and lift the embedding to $j:V[G]\to M[j(G)]$. Let $s=h(j''\mu)$, which covers $j''\mu$ and therefore generates a $\mu$-strong compactness measure in $V[G]$. So $\kappa$ is strongly compact in $V[G]$ and the least measurable cardinal.

Now, with regard to your question, the point is that $|s|=h(\mu)$, which is equal to $\mu$ in our case because $\mu$ is regular and above $\kappa^+$. So the lifted embedding $j:V[G]\to M[j(G)]$ has your small cover property, in a situation where $\kappa$ is the least measurable cardinal. And as you pointed out, we even have that the order type of the covering set $s$ is $\mu$. So in general we cannot get any degree of supercompactness from this small cover property.

I am unsure at the moment about the dual situation, where one has a strongly compact cardinal that has only large covers. Perhaps one can get this at least for $\mu=\kappa^+$ in the above model, since in that case $h(\mu)$ is larger than $\mu$.