Definitions of an additive functor In my Rings and Modules class, we defined additive functors this way:

A (covariant/contravariant) functor $F$ is called additive if:

(i). $F0 = 0$, where $0$ is the zero-module.

(ii). For any modules $M, N$ we have $F(M \oplus N) = FM \oplus FN$.

I looked around online and I saw this definitions, which looks the same but uses morphisms instead of modules:

A (covariant/contravariant) functor $F$ is called additive if:

(i). $F0 = 0$, where $0$ is the zero-map.

(ii). For any modules $M, N$ and morphisms $\phi, \psi: M \longrightarrow N$ we have $F(\phi + \psi) = F \phi + F \psi : FM \longrightarrow FN$.

Now it seems clear, intuitively, how these two concepts are equivalent. However, I always have a lot of trouble proving elementary stuff with functors, because I always think if what I'm doing is allowed.

I was trying to prove that the first two conditions imply the other two. So I tried to rephrase them in terms of compositions of homomorphisms.

(i). Let $\phi : M \longrightarrow N$ be the zero-map. Then $\phi$ is the same as $\iota \circ \phi_0$, where $\phi_0 : M \longrightarrow 0$ and $\iota : 0 \longrightarrow N$. Then $F \phi = F(\iota \circ \phi_0) = F \iota \circ F \phi_0$ is the zero-map, because we are assuming that $F0 = 0$.

(ii). Let $\phi, \psi: M \longrightarrow N$ be homomorphisms. Then $\phi + \psi$ is the same as the following sequence:
$$ M \overset{id_M \oplus id_M}{\longrightarrow} M \oplus M \overset{(\phi, \psi)}{\longrightarrow} N \oplus N \overset{id_N + id_N}{\longrightarrow}N.$$
At this point, however I am stuck. I don't know how to justify that applying $F$ to this sequence gives $F\phi + F\psi$, since natural stuff such as inclusions and projections are not necessarily preserved. How should I go on?

I haven't yet tried proving the other direction...
Your first definition is imprecise. What you should say is that the universal property of coproducts furnishes a natural map $F(M) \oplus F(N) \to F(M \oplus N)$ and you want these natural maps to be isomorphisms. Anyway, see this blog post for a thorough discussion. The upshot is that in an additive category direct sums determine the abelian group structure on morphisms.
To show the first definition implies the second, see The Stacks Project, Lemma 12.7.1.
To show the second fefinition implies the first, see The Stacks Project, Lemma 12.3.7. Note that now products equal to coproducts by The Stacks Project, Lemma 12.3.4. The key here is to use an equivalent definition of binary (co)products. Namely, as in The Stacks Project, Remark 12.3.6, if $f:x\to z$, $j:y\to z$, $p:z\to x$ and $q:z\to y$ satisfy $p\circ i=id_{x}$, $q\circ j=id_{y}$, $p\circ j=0$, $q\circ i=0$ and $i\circ p + j\circ q=id_{z}$, then $z$ with these maps is the (co)product of $x$ and $y$.
To see this, we verify for being the product. For any $f:w\to x$ and $g:w\to y$, if there is some $h:w\to z$ s.t. $p\circ h=f$ and $q\circ h=g$, then we have $$ i\circ f+j\circ g=(i\circ p)\circ h+(j\circ q)\circ h= (i\circ p + j\circ q)\circ h = h. $$ And $i\circ f+j\circ g$ can be such $h$ since $$p\circ (i\circ f+j\circ g)=(p\circ i)\circ f +(p\circ j)\circ g= f+0=f $$ and also $q\circ(i\circ f+j\circ g)=g $. So there is a unique morphism from $w$ to $z$ as desired. As for being the coproduct, the proof is similar. Then we see that being additive on the hom sets can preserve these equalities of morphisms. So $F$ will send the binary (co)product to a binary (co)product.
As for the zero object, see the third characterization in The Stacks Project, Lemma 12.3.2 and use the same reasoning.