Toy examples for Kan extensions
If $J$ is only locally small, the definition of $\mathrm{Lan}(f)(F)$ still makes sense even if $f^\ast$ is not definable without jumping to a bigger universe. It satisfies the 'local' adjointness property : there is a natural transformation $\eta \colon F \to \mathrm{Lan}(f)(F) \circ f$ universal in the sense that any $F \to G \circ f$ factors through $\eta$.
In this framework, it is very useful to consider the left Kan extension of a functor $F$ along the Yoneda embedding. For example, if you consider the standard cosimplicial space $\Delta^\bullet \colon \boldsymbol\Delta \to \mathsf{Top}$, then its left Kan extension along the Yoneda embedding is precisely the geometric realization functor $|\!-\!| \colon \hat{\boldsymbol\Delta} \to \mathsf{Top}$.
Still with simplicial sets, the left Kan extension of the full inclusion $i\colon \boldsymbol\Delta \to \mathsf{Cat}$ is the fundamental category functor $\tau_1 \colon \hat{\boldsymbol\Delta} \to \mathsf{Cat}$ (that is the functor mapping $X$ to the category whose objects are the elements of $X_0$ and whose morphims are freely generated by $X_1$ under the relation of composition given by the three face maps $X_2 \to X_1$).
In the two previous examples, $\eta$ is actually $\mathrm{id}_F$ (I mean the lax commutative triangle is actually commutative) and the functor $\mathrm{Lan}(f)(F)$ admits a right adjoint (the singular functor in the first case and the nerve functor in the second case). More generally, any functor $F \colon \mathcal I \to \mathcal C$ from a small category $\mathcal I$ to a cocomplete category $\mathcal C$ admits a left Kan extension $\mathrm{Lan}(\mathfrak h^\mathcal I)(F)$ along the Yoneda embedding $\mathfrak h^\mathcal I \colon \mathcal I \to \hat{\mathcal I}$ with structural map $\eta = \mathrm{id}_F$. Moreover this left Kan extension admits $c \mapsto \hom_{\mathcal C}(F-,c)$ as a right adjoint.
(I hope I didn't mess any of these up.) EDIT: I did mess a couple up, the seond and third, many thanks to @JoshuaMeyers for catching the mistakes! They have been corrected.
For $f : B\mathbb{N} \to B\mathbb{Z}$ (where $BM$ is the one-object category corresponding to the monoid $M$) given by the inclusion, $\mathrm{Lan}(f)$ sends an endomorphism $g:X \to X$ to the endomorphism of $Y := \mathrm{colim}(X \xrightarrow{g} X \xrightarrow{g} \cdots)$ that is given by $g$ on every copy of $X$. This is the usual construction for "inverting $g$".
Let $I = \{0 \to 1\}$. For $f : I \to B \mathbb{N}$, the morphism that picks out of the generator of $\mathbb{N}$, $\mathrm{Lan}(f)$ sends a morphism $g : X \to Y$ to the endomorphism on $X \sqcup \coprod_\mathbb{N} Y$ mapping $X$ to the first summand $Y$ by means of $g$, and sending each summand $Y$ to the next via the identity.
Let $I$ be as above and $J$ be the category with two objects and a unique isomorphism between them. For the inclusion $f : I \to J$, again $\mathrm{Lan}(f)$ sends a morphism $g: X \to Y$ to the identity on $Y$ (ignoring $X$ and $g$).
Let $P = I \coprod_{\{0,1\}} I$ be the pair of parallel arrows. For the fold map $f : P \to I$, $\mathrm{Lan}(f)$ sends a pair $g, h : X \to Y$ to the canonical morphism $X \to \mathrm{coeq}(g,h)$ (note the domain is $X$, not $Y$).
(Before you scold me for this one, consider that one person's "proposition" is another's "family of examples".) If $f : C \to D$ is an opfibration, then $\mathrm{Lan}(f)$ is computed by taking colimits over the fibres: it sends a functor $g : C \to E$ to $d \mapsto \mathrm{colim}(g|_{f^{-1}(d)})$, where $f^{-1}(d)$ is category consisting of the objects of $C$ mapping to $d \in D$ and those morphisms between them that map to the identity on $d$.
Here's one you might know. If $f: H \to G$ is a group homomorphism, then $f^*: [G,\mathsf{Vect}] \to [H,\mathsf{Vect}]$ is restriction of group representations, denoted $\operatorname{Res}_f$. The left adjoint $\operatorname{Ind}_f: [H,\mathsf{Vect}] \to [G,\mathsf{Vect}]$ is the induced representation functor. If $G$ and $H$ are finite, then $\operatorname{Ind}_f$ is also right adjoint to $\operatorname{Res}_f$. Induced representations can be written down in an explicit formula which really comes from the general formula you give, but in some ways feels more concrete.