It is not true that one can obtain the coarse moduli space of a functor or stack $F$ by sheafifying $F$ or $\pi_0(F)$, so probably the answer to both of your questions is NO.

I'll take the following definition of a coarse moduli space of a functor or stack $F: (Sch)^{op} \rightarrow \mathrm{Gpoid}$: It's an algebraic space $X$ together with a natural transformation $F \rightarrow X$ such that

1. $X$ is universal for maps from $F$ to algebraic spaces (a bit stronger than being universal to schemes), and

2. For any algebraically closed field $k$, the natural map $F(k) \rightarrow X(k)$ is a bijection.

For this definition (which I think is the standard one), the coarse moduli space is typically NOT obtained by sheafifying $\pi_0(F)$. For example, take $F$ to be the Deligne-Mumford stack $[\mathbb{A}^1/(\mathbb{Z}/2)]$, for $\mathbb{A}^1 = \mathrm{Spec \ }\mathbb{C}[x]$ and $\mathbb{Z}/2$ acting in the obvious way by $x \mapsto -x$. Then the coarse moduli space of $F$ is given by $X = \mathbb{A}^1$, and the map $F \rightarrow X$ is induced by the map $x \mapsto x^2$ from $\mathbb{A}^1$ to $\mathbb{A}^1$. But $\mathbb{A}^1$ is not isomorphic to the sheafification of $\pi_0(F)$ for the etale topology (and probably not for any subcanonical topology, although I'm not sure how to prove this at the moment).

To see why, let $Y$ be the etale sheafification of $\pi_0(F)$, and let $T_0 F$, $T_0 X$, and $T_0 Y$ be the tangent spaces to these functors at 0 (by which I mean the set of lifts of $0 \in F(\mathbb{C})$ to $F(\mathbb{C}[\epsilon])$, where $\epsilon^2 = 0$). If $X$ were given by the etale sheafification of $\pi_0(F)$, there would be a natural map $Y \rightarrow X$ induced from the projection $F \rightarrow X$ which would be an isomorphism on tangent spaces. But one sees that the map $T_0 F \rightarrow T_0 X$ is the zero map since this is true for the map $x \mapsto x^2: \mathbb{A}^1 \rightarrow \mathbb{A}^1$. On the other hand, the map $T_0 X \rightarrow T_0 Y$ is non-zero, because $\mathbb{C}[\epsilon]$ has no non-trivial etale covers and so $T_0 Y = T_0(\pi_0(F))$, which is isomorphic to $\mathbb{C}/(\mathbb{Z}/2)$ with the projection map $T_0 F \rightarrow T_0 Y$ identified with the quotient map $\mathbb{C} \rightarrow \mathbb{C}/(\mathbb{Z}/2)$.

If you look at the proof of the existence of coarse moduli spaces, you'll see that it's done by first considering the case of quotienting an affine scheme by a finite flat groupoid, where the coarse moduli space is given by Spec of the invariants. One then carefully glues together the coarse moduli spaces on open patches. This construction makes me suspect that there is no direct (i.e., global) construction of the coarse moduli space, at least not one that is easy to work with.

It is not true that one can obtain the coarse moduli space of a functor or stack $F$ by sheafifying $F$ or $\pi_0(F)$, so probably the answer to both of your questions is NO.

I'll take the following definition of a coarse moduli space of a functor or stack $F: (Sch)^{op} \rightarrow \mathrm{Gpoid}$: It's an algebraic space $X$ together with a natural transformation $F \rightarrow X$ such that

1. $X$ is universal for maps from $F$ to algebraic spaces (a bit stronger than being universal to schemes), and

2. For any algebraically closed field $k$, the natural map $F(k) \rightarrow X(k)$ is a bijection.

For this definition (which I think is the standard one), the coarse moduli space is typically NOT obtained by sheafifying $\pi_0(F)$. For example, take $F$ to be the Deligne-Mumford stack $[\mathbb{A}^1/(\mathbb{Z}/2)]$, for $\mathbb{A}^1 = \mathrm{Spec \ }\mathbb{C}[x]$ and $\mathbb{Z}/2$ acting in the obvious way by $x \mapsto -x$. Then the coarse moduli space of $F$ is given by $X = \mathbb{A}^1$, and the map $F \rightarrow X$ is induced by the map $x \mapsto x^2$ from $\mathbb{A}^1$ to $\mathbb{A}^1$. But $\mathbb{A}^1$ is not isomorphic to the sheafification of $\pi_0(F)$ for the etale topology (and probably not for any subcanonical topology, although I'm not sure how to prove this at the moment).

To see why, let $Y$ be the etale sheafification of $\pi_0(F)$, and let $T_0 F$, $T_0 X$, and $T_0 Y$ be the tangent spaces to these functors at 0 (by which I mean the set of lifts of $0 \in F(\mathbb{C})$ to $F(\mathbb{C}[\epsilon])$, where $\epsilon^2 = 0$). If $X$ were given by the etale sheafification of $\pi_0(F)$, there would be a natural map $Y \rightarrow X$ (induced from the projection $F \rightarrow X$) which would be an isomorphism on tangent spaces. But one sees that the map $T_0 F \rightarrow T_0 X$ is the zero map since this is true for the map $x \mapsto x^2: \mathbb{A}^1 \rightarrow \mathbb{A}^1$. On the other hand, the map $T_0 X \rightarrow T_0 Y$ is non-zero, because $\mathbb{C}[\epsilon]$ has no non-trivial etale covers and so $T_0 Y = T_0(\pi_0(F))$, which is isomorphic to $\mathbb{C}/(\mathbb{Z}/2)$ with the projection map $T_0 F \rightarrow T_0 Y$ identified with the quotient map $\mathbb{C} \rightarrow \mathbb{C}/(\mathbb{Z}/2)$.

If you look at the proof of the existence of coarse moduli spaces, you'll see that it's done by first considering the case of quotienting an affine scheme by a finite flat groupoid, where the coarse moduli space is given by Spec of the invariants. One then carefully glues together the coarse moduli spaces on open patches. This construction makes me suspect that there is no direct (i.e., global) construction of the coarse moduli space, at least not one that is easy to work with.