I think that Aurelien Djament's answer is essentially correct, but I want to nitpick a bit.

 1. If $\mathcal A$ is any locally finitely presentable category and $\mathcal C \subseteq \mathcal A$ is any strong generator of finitely-presentable objects, then every finitely-presentable object $X \in \mathcal A$ lies in the closure of $\mathcal C$ under finite colimits. So $X$ is a finite colimit of finite colimits of ... of finite colimits of objects of $\mathcal C$ -- an "$n$-fold" finite colimit of objects of $\mathcal C$. But $X$ need not be a "1-step" finite colimit of objects of $\mathcal C$. For example, I don't think every finitely-presented group is a finite colimit of copies of $\mathbb Z$.

 2. One might strengthen the hypotheses and ask: if $\mathcal A$ is a locally finitely presentable category and $\mathcal C \subseteq \mathcal A$ is a _dense_ generator, then is every finitely-presentable object $X \in \mathcal C$ a finite colimit of objects of $\mathcal C$? I don't know the answer to this.

 3. But let's focus on the question at hand, i.e. the case where $\mathcal A = \hat {\mathcal C}$ is a presheaf category and $\mathcal C$ is the representables. Let $\tilde {\mathcal C}$ comprise the finite colimits of representables. Then indeed, $\tilde {\mathcal C}$ is closed under finite colimits. This is clear for finite coproducts -- just take the coproduct of the indexing diagrams for the colimits. Now let $A\rightrightarrows B \to C$ be a coequalizer where $A,B \in \tilde {\mathcal C}$. Then there is an epimorphism $\amalg_i X_i \to A$ and a coequalizer diagram $\amalg_j Y_j \rightrightarrows \amalg_k Z_k \to B$ where $X_i,Y_j,Z_k \in \mathcal C$ and the coproducts are finite. The composite maps $\amalg_i X_i \to A \rightrightarrows B$ lift to maps $\amalg_i X_i \rightrightarrows \amalg_k Z_k$. Then we have that $C$ is the coequalizer of the two induced maps $(\amalg_i X_i) \amalg (\amalg_j Y_j) \rightrightarrows \amalg_k Z_k$.

I want to emphasize that here we heavily used the fact that we're in a presheaf category.

 4. I agree that any category which has finite colimits and filtered colimits has all colimits. But Aurelian's second bullet seems to suggest something stronger -- that if $X$ is a colimit of objects of $\mathcal C$, then $X$ is a filtered colimit of finite colimits of objects of $\mathcal C$. I don't have a counterexample, but I'm not sure this is true. The closest I can convince myself of is that $X$ is a coequalizer of coproducts of objects of $\mathcal C$, and therefore a coequalizer of filtered colimits of finite coproducts of objects of $\mathcal C$ -- but this only ensures that $X$ is a finite colimit of filtered colimits of finite colimits of objects of $\mathcal C$.

 5. But using (3), Aurelian's third bullet goes through with some modification. As in any locally finitely presentable category $\mathcal A$ with strong generator $\mathcal C$, any finitely-presentable object is in the closure of the $\mathcal C$ under finite colimits. By (3), in the case $\mathcal A = \hat{\mathcal C}$, the closure of $\mathcal C$ under finite colimits consists exactly of $\tilde{\mathcal C}$, the objects which are "1-step" finite colimits of representables. Here, (3) is actually used in 2 places: first to ensure that the category $\tilde C \downarrow X$ is filtered (this being the diagram which indexes the canonical colimit for $X$), and second to ensure that $\tilde{\mathcal C}$ is closed under retracts.