Concerning (i):

If $F\colon\mathcal{F}(N,\mathbb R)\to \mathcal{F}(M,\mathbb R)$ is of the form $F_T$, $F$ is a homomorphism of algebras, this is an obvious necessary condition. It turns out, it is also a sufficient one.

There is a contravariant functor $\mathcal A$ from the category of smooth manifolds and smooth maps to the category of algebras, which send every manifold $M$ to the algebra of smooth functions on it:

$$ \mathcal A\colon M\mapsto \mathcal{F}(M,\mathbb R) $$ and smooth maps to the corresponding algebra homomorphisms:

$$ \mathcal A\colon \{T\colon M\to N\}\mapsto\{F_T\}. $$

Your question is what is the image of this functor on the level of $\hom$'s. The answer is:

$\mathcal A$ is full and faithful, in particular, any homomorphism of algebras $F\colon\mathcal{F}(N,\mathbb R)\to \mathcal{F}(M,\mathbb R)$ is of the form $F_T$.

The proof can be found in http://www.emis.de/monographs/KSM/kmsbookh.pd - see 35.9 and 35.10 therein.

Concerning (i):

If $F\colon\mathcal{F}(N,\mathbb R)\to \mathcal{F}(M,\mathbb R)$ is of the form $F_T$, $F$ is a homomorphism of algebras, this is an obvious necessary condition. It turns out, it is also a sufficient one.

There is a contravariant functor $\mathcal A$ from the category of smooth manifolds and smooth maps to the category of algebras, which send every manifold $M$ to the algebra of smooth functions on it:

$$ \mathcal A\colon M\mapsto \mathcal{F}(M,\mathbb R) $$ and smooth maps to the corresponding algebra homomorphisms:

$$ \mathcal A\colon \{T\colon M\to N\}\mapsto\{F_T\}. $$

Your question is what is the image of this functor on the level of $\hom$'s. The answer is:

$\mathcal A$ is full and faithful, in particular, any homomorphism of algebras $F\colon\mathcal{F}(N,\mathbb R)\to \mathcal{F}(M,\mathbb R)$ is of the form $F_T$.

The proof can be found in http://www.emis.de/monographs/KSM/kmsbookh.pdf - see 35.9 and 35.10 therein.