Rif to Shawn Hedman, A first course in logic : An Introduction to Model Theory, Proof Theory, Computability, and Complexity (2004), page 80.
I think there are some misprints ...
Literals are defined for propositional calculus [page 28] :
Definition 1.51 A literal is an atomic formula or the negation of an atomic
For first order logic, see the Definition 2.1 of atomic formula [page 55].
Referring to Example 2.51, page 80, they are graphs [see page 66]; thus, the only atomic formulae are like $R(x,y)$ where the relation $R$ express the fact that vertices $x,y$ are adjacent.
According to :
Definition 2.49 Let $\mathcal V$ be a vocabulary and let $M, N$ be $\mathcal V$-structures. A function $f : M \to N$ preserves the $\mathcal V$-formula $\varphi(\overline x)$ if, for each tuple $\overline a$ of elements in $M$, $M \vDash \varphi(\overline a)$ implies $N \vDash \varphi(f(\overline a))$.
In the example above, we have to correct two misprints : in the formulae defining $g : M \to N$ we have to replace $f$ with $g$ and the final statement must be : Then $f$ is a literal embedding and $g$ is not.
You can check that if $M \vDash R(a_1,a_2)$, then $N \vDash R(f(a_1), f(a_2))$, for any $a_1,a_2 \in M$.
For $g$ instead, we have that $R(B,C)$ holds in $M$, while $R(e,d)=R(g(B), g(C))$ does not hold in $N$, because $B,C$ are adjacent in graph $M$ while $e,d$ are not adjacent in $N$.
For the relation between the three concepts in issue, consider Example 2.52 :
Let $id : \mathbb N_< \to \mathbb Z_<$ be the identity function defined by $id(x) = x$. This is a literal embedding.
In the language with the only binary predicate $<$, the atomic formulae are like $x < y$, and it is true that if $\mathbb N_< \vDash n < m$, then $\mathbb Z_< \vDash n < m$.
Since $\mathbb N_< \vDash ¬∃x(x < 0)$ and $\mathbb Z_< \vDash ∃x(x < 0)$ this embedding
does not preserve the formula $¬∃x(x < y)$, and so it is not an elementary
embedding [in $\mathbb N$ there are no numbers less than $0$, while in $\mathbb Z$ we have that $-1 < 0$].
The identity function from $\mathbb Q_<$ to $\mathbb R_<$, on the other hand, is an elementary embedding [this will be proved in Chapter 5]
but it is not an isomorphism, because the two structure have not the same cardinality.