What is a simple definition of the pullback of a section?

There is a pretty geometric answer, if $X, Y$ are schemes and $G = \mathcal{O}_Y$ is the structure sheaf. In that case, a global section $s \in \Gamma(Y, \mathcal{O}_Y)$ is the same as a morphism $Y \to \mathbb{A}^1_\mathbb{Z}$. The pull-back $f^*s$ then corresponds to the composed morphism $$X \xrightarrow{f} Y \to \mathbb{A}^1_\mathbb{Z},$$ and defines in that way a global section $f^*s \in \Gamma(X, \mathcal{O}_X)$.

For abritrary ringed spaces $X,Y$, and sheaf $G$, verify that a global section $s \in \Gamma(Y, G)$ is the same as a $\mathcal{O}_Y$-module homomorphism $s: \mathcal{O}_Y \to G$. Pulling-back that morphism yields $$f^*s: \mathcal{O}_X = f^*\mathcal{O}_Y \to f^*G,$$ which defines the global section $f^*s \in \Gamma(X, f^*G)$. Here it is important to verify, that $f^*$ is a functor $\operatorname{Mod}(Y) \to \operatorname{Mod}(X)$.

Also verify that this yields the first construction in the case $G = \mathcal{O}_Y$.

red_trumpet's answer is quite categorical, choosing to talk about morphisms and apply functoriality in place of talking about elements as much as possible. This is a good perspective, but it can also be useful to understand what's going on directly at the level of elements, so I thought I'd add another answer. In fact, there's a fairly satisfying "only thing it could be" definition of $f^*s$.

Suppose we have: $f\colon X\to Y$ a morphism of schemes, $G$ a sheaf of $\mathcal{O}_Y$-modules, $U\subseteq Y$ open, and $s\in G(U)$.

First of all, what should $f^*s$ be? The most natural answer is that we should have $f^*s \in f^*G(f^{-1}U)$. Ok, now let's unpack the definitions. Set $V = f^{-1}(U)$.

First, $f^*G = f^{-1}G\otimes_{f^{-1}\mathcal{O}_Y}\mathcal{O}_X$. This tensor product is the sheafification of the tensor product presheaf $f^{-1}G\otimes^{\text{p}}_{f^{-1}\mathcal{O}_Y}\mathcal{O}_X$ (that p is for "presheaf"). And there is a map of presheaves from any presheaf to its sheafification. So to give an element of $f^*G(V)$, it suffices to give an element of $$(f^{-1}G\otimes_{f^{-1}\mathcal{O}_Y} ^{\text{p}}\mathcal{O}_X)(V) = f^{-1}G(V)\otimes_{f^{-1}\mathcal{O}_Y(V)}\mathcal{O}_X(V).$$ Now given an $R$-module $M$ and an extension of rings $R\to S$, there is a natural map $m\mapsto m\otimes_R 1_S$ from $M$ to the "base changed" $S$-module $M\otimes_R S$ So to give an element of the module above, it suffices to give an element of the module $f^{-1}G(V)$.

The sheaf $f^{-1}G$ is the sheafification of the presheaf $f^{-1}_\text{p}G$ defined by $$f^{-1}_\text{p}G(V) = \text{colim}_{W\supseteq f(V)} G(W).$$ Again, there is a map from any presheaf to its sheafification, so to give an element of $f^{-1}G(V)$, it suffices to give an element of $f^{-1}_\text{p}G(V)$.

We have $U\supseteq f(f^{-1}U) = f(V)$, so $G(U)$ is one of the modules included in the colimit diagram. That is, we have a map $$G(U) \to \text{colim}_{W\supseteq f(V)} G(W) = f^{-1}_\text{p}G(V).$$

The image of $s\in G(U)$ under this long chain of maps $$G(U)\to f^{-1}_\text{p}G(V) \to f^{-1}G(V) \to f^{-1}G(V)\otimes_{f^{-1}\mathcal{O}_Y(V)} \mathcal{O}_X(V)\to f^*G(V)$$ is the desired element $f^*s\in f^*G(V)$.

We treat pullbacks of global sections, nothing changes if we restrict to some other open set. We will be using the fact that tensor product of sheaves commutes with restriction to open sets.

Call $f: X \rightarrow Y$ our morphism of schemes. Let $\mathscr{G}$ be an $\mathcal{O}_Y$-module. There is a canonical morphism $\mathscr{G} \rightarrow f_* f^* \mathscr{G}$ of $\mathcal{O}_Y$-modules. This induces a morphism of global sections $\psi:\Gamma(Y,\mathscr{G}) \rightarrow \Gamma(X,f^*\mathscr{G})$. The question is how to describe concretely the image of $s \in \Gamma(Y,\mathscr{G})$ under this map.

It is instructive to begin with the special case $\mathscr{G}=\mathcal{O}_Y$. Consider the following rings $A=\Gamma(Y,\mathcal{O}_Y), \, B = \Gamma(X,f^{-1} \mathcal{O}_Y), \, C = \Gamma(X,\mathcal{O}_X)$. The data of $f$ include a morphism $\mathcal{O}_Y \rightarrow f_* \mathcal{O}_X$. This induces a ring homomorphism $\phi_1: A \rightarrow C$ on global sections. Now the canonical morphism $\mathcal{O}_Y \rightarrow f_* f^{*} \mathcal{O}_Y$ induces a ring homomorphism $\phi_1':A \rightarrow \Gamma(X,f^* \mathscr{G})$. The key is that $\phi_1$ and $\phi_1'$ have the same image. To see this note that we have a ring homomorphism $\phi_2: A \rightarrow B$ induced by $\mathcal{O}_Y \rightarrow f_* f^{-1} \mathcal{O}_Y$. We also have a $\phi_3 : B \rightarrow C$ induced by $f^{-1} \mathcal{O}_Y \rightarrow f^{-1} f_* \mathcal{O}_X \rightarrow \mathcal{O}_X$. Moreover $\phi_1 = \phi_3 \circ \phi_2$. Now $\Gamma(X,f^* \mathcal{O}_Y) = B \otimes_B C$. Finally by checking the definition of $\phi_1'$ we can see that $\phi_1'$ sends $s$ to $\phi_2(s) \otimes 1 \in B \otimes_B C$ which we can identify with $\phi_3(\phi_2(s))=\phi_1(s)$.

For general $\mathscr{G}$ the picture is analogous with the note that $\Gamma(X,f^*\mathscr{G}) = \Gamma(X,f^{-1}\mathscr{G}) \otimes_B C$. That is $\psi$ sends global sections of $\mathscr{G}$ to their canonical image in the $B$-module $\Gamma(X,f^{-1} \mathscr{G})$ and then to their image in the $C$-module $\Gamma(X,f^{-1} \mathscr{G}) \otimes_B C$, the latter being an extended module by scalars according to $\phi_3: B \rightarrow C$.