Here is a motivation for hocolims, for concreteness look at pushouts of topological spaces:

The usual pushout functor $colim: Top^P \rightarrow Top$, where $P:=\bullet \leftarrow \bullet \rightarrow \bullet$ is the diagram of a pushout datum, does not respect weak equivalences:
Take for example the diagrams $pt \leftarrow S^1 \rightarrow pt$ and $D^2 \leftarrow S^1 \rightarrow D^2$ in $Top$. Since $D^2$ (the 2-dimensional disk) is contractible, there is a map between the two diagrams consisting of weak equivalences, i.e. a weak equivalence in $Top^P$. But if we apply colim to both diagrams we get non-equivalent objects in $Top$, namely $pt$ in the first case and $S^2$ in the second.

Thus the pushout functor $colim: Top^P \rightarrow Top$ does not take weak equivalences to weak equivalences and we can not invoke the universal property of $Ho(Top^P)$ to get a functor $Ho(colim):Ho(Top^P) \rightarrow Ho(Top)$ making the square commute which has $colim$ on top, $Ho(colim)$ at the bottom and the projections to the homotopy categories at the left and right. The best we can do is to find the terminal functor "$hocolim$" which allows filling this square with a natural transformation - which is exactly the Kan extension property you cited.

For model categories (but not only there, e.g. also Baues cofibration categories) you can define that functor via the top level, i.e. going $Top^P \rightarrow Top$, in a particularly neat way. Back to the example: Topologists noticed that the colim functor, when restricted to pushout data $A \leftarrow B \rightarrow C$ in $Top$ with $B$ cofibrant and the arrows cofibrations, *does* preserve weak equivalences. E.g. building the pushout of two inclusions of $S^1$ into contractible spaces such that there is "enough space" around the image of $S^1$ inside those spaces, you will always get something equivalent to $S^2$ - try it out! So the recipe for computing the $hocolim$ is first replacing your diagram by one with these properties (cofibrant replacement - this is an endofunctor which preserves weak equivalences) and then apply $colim$ - this does now preserve weak equivalences and thus descends to a functor between the homotopy categories.

So the intuition about $hocolim$ - which is good in great generality - is that it is the best approximation to $colim$ which preserves weak equivalences ("is homotopy invariant"); the cofibrant replacement construction stems from the fact that this is a class of objects where $colim$ is already homotopy invariant.

The story for homotopy limits is of course dual, instructive examples are homotopy fibers and homotopy fixed point objects.

**Edit:** Here is how you see that $colim \circ Q$ is $hocolim$ - assuming that we have a cofibrant replacement *functor* $Q$ on $Top^P$: you can simply check the universal property. So let $F:Ho(Top^P) \rightarrow Ho(Top)$ be a functor and $\tau:F \circ Ho_{Top^P} \rightarrow Ho_{Top} \circ colim$ a natural transformation. Here the functors $Ho_*$ are the projections to the homotopy categories which leave objects unchanged and map morphisms to their homotopy classes - I will omit them from the notation from now on; so we consider a natural transformation $\tau: F \rightarrow colim$ and have to show that it factors through $colim \circ Q$.

The cofibrant replacement functor $Q$ comes with a natural weak equivalence $Q \rightarrow id$. Composing with $F$ gives a natural isomorphism $F \circ Q \rightarrow F$. Now for each pushout datum $D \in Top^P$ we have the chain

$$F(D) \leftarrow F(QD) \rightarrow colim(QD) \rightarrow colim(D)$$

where the middle arrow is $\tau_{QD}$ (the natural transformation $\tau$ at the object $QD$) and the the outer two arrows arise by applying $F,colim$ respectively to $QD \rightarrow D$. The left arrow can be gone backwards because it is an isomorphism. The whole way from left to right is then equal to $\tau_D$ because of the naturality of $\tau$ (flip the outer arrows downwards, fill in $\tau_D$ below and you got the naturality square). This shows that each $\tau_D$ factors through $colim \circ Q$. To see that this factorization is natural in $D$, observe that $QD \rightarrow D$ and $\tau_{Q-}$ are natural in $D$.

**Edit2:** While Harry generously granted me a check mark after the above, Tom Goodwillie is of course right that a statement about uniqueness is in order. Here is why the above factorization is unique *at the level of homotopy categories*: Given any factorization $\tau=i \circ \tau'$ of our given $\tau$, for cofibrant $D$ it will factorize as

$$\tau_D=i_D \circ \tau_D^':F(D) \rightarrow colim(QD) \rightarrow colim(D)$$

where the second arrow $i_D$ is an isomorphism, because it is $colim$ of a weak equivalence between cofibrant objects. So the first factor must be $\tau_D^'=i^{-1}_D \circ \tau_D$, there is no choice here. For general $D$ we can express $\tau_D^'$ as

$$F(D) \leftarrow F(QD) \rightarrow colim(QQD) \rightarrow colim(QD)$$ by walking around the naturality square for $\tau'$ for the morphism $QD \rightarrow D$. So we also have no choice for non-cofibrant $D$.

**Edit3**: As spotted by Tom Goodwillie I implicitly (and unconsciously!) used here that $i_{QD}=Q(i_D)$ - see the comments. You can prove this by drawing all available naturality diagrams with $QQQD$ at the left upper corner, seeing from those first that $QQ(i_D)=Q(i_{QD})=i_{QQD}$ and that hence $i_{QD} \circ QQ(i_D) = Q(i_D) \circ i_{QQD} = Q(i_D) \circ Q(i_{QD}) = Q(i_D) \circ QQ(i_D)$. Now one can cancel the isomorphism $QQ(i_D)$ on both sides.

Now I switch the notation back to mentioning the $Ho_*$: This showed that $Ho(colim \circ Q)$ is the Kan extension of $Ho_{Top} \circ colim$ along $Ho_{Top^P}$. That is already good to know, but it was just level zero of showing that $colim \circ Q:Top^P \rightarrow Top$ is the *homotopy terminal* homotopy invariant functor. To proceed you have to produce a simplicial set; e.g. do your localisation by passing to some model of an $(\infty, 1)$-category (e.g. hammock localisation, coherent nerve, ...), or use framings, or consider the nerve of some category of functors into $colim \circ Q$ and then show that a certain space is contractible. This should be done at a place where you can draw pictures, unlike in a MathOverflow answer.