Intuitive understanding of a non-full functor A functor $T : C \to B$ is full if for each $c,c' \in C$ and for each $g : T c \to T c'$,  there is at least a morphism $f : c \to c'$ such that $g = T f$.
What does this tell me about non-full functors?
My understanding is that, as soon as there's a morphism $g : T c \to T c'$ in $B$, this morphism must come from a morphism in $C$; on the other hand, if the functor is not full, i.e. e.g. there is no morphism $c \to c'$ in $C$, then there  must be another pair of objects $d,d' \in C$ such that

*

*$T d = T c$,

*$T d' = T c'$,

*$d \neq c \vee d' \neq c'$,

*$\exists h:d \to d'$ such that $T h = g$.

Is this correct?
 A: Why should there be such a pair ? It is completely possible that $h$ is simply not in the image of the functor $T$. (*)
"Most" functors aren't full, this is the "generic" situation, full functors are something special, that's it.
There's a misconception in your question "as soon as there's a morphism in $B$, it must come from $C$" : that is simply not true.
Let's look at an example, one of the most classical ones (otherwise it's easy to cook up counterexamples by hand, this one is "found in nature"): the forgetful functor $U: Grp\to Set$ from the category of group and group morphisms to the category of sets and functions between them.
This is definitely not full : given $G,H$ groups, there are tons of applications $UG\to UH$ that aren't group morphisms.
Ok, that doesn't tell me that there can't be $G',H'$ with $UG' = UG, UH' = UH$ for any application. Well, it does : here's a set theoretic condition that any group morphism must satisfy: if $f:G\to H$ is a morphims of groups then for any $x,y \in H, f^{-1}(x)$ and $f^{-1}(y)$ either have the same cardinality, or one of them is empty.
Indeed, if $x\in f(G)$ then $f^{-1}(x)$ has the cardinality of $\ker(f)$. In particular, if $h: UG\to UH$ does not satisfy this property, there is no group structure on $UG,UH$ that can "lift" $h$.
So the definition of full functors tells you nothing about non-full ones, it just tells you something about full functors.
A: From the comments:
Saying that a functor $F : C \to D$ is not full is saying that there are some objects $Fx$ and $Fy$ in $D$ that have ~bonus arrows~ between them that don't come from arrows in $C$.
For instance, look at the category with 2 objects and one arrow, shown below:
$$x \to y$$
Then a functor from $C \to \mathsf{Set}$, say, is a choice of 2 sets $X$ and $Y$, plus a choice of function $f : X \to Y$.
As long as there are multiple choices of $f$, this functor cannot be full.
More generally, saying that a functor is full is saying that we can understand every arrow in $D(Fx,Fy)$ as the image of an arrow that already existed in $C(x,y)$. Saying that a functor is not full is saying that there are some extra arrows between $Fx$ and $Fy$ that exist in $D$ but not $C$.

I hope this helps ^_^
