The analogous of a module $M$ over a ring $A$ is a (quasi coherent) sheaf of modules over a scheme $X$. Indeed, in the case $X=\operatorname{Spec} A$ is an affine variety, the two notions coincide. When
First point: how one can think of a sheaf $F$ over $X$? Well, you can take the associated étalé space and think of it as a space $p: E \to X$, such that
$$F(U) \simeq \{e \in E: p(e) \in U\}$$
Note that sections are $R$-modules in this case. In other words, you can always think a sheaf as a sheaf of sections. A cool way to think about sections are a sort of "generalized functions" that can take more than one value. For example, take the function $x \mapsto x^2$ over $\mathbb{C}$: the local sections in a neighborhood of 1 are exactly the two functions $\sqrt{x}, -\sqrt{x}$.
Another set of geometric examples of sheaf - despite less general and less common in algebraic geometry - is the sheaf of solutions to a given constraint (as a differential equation) : locally it can have a lot of solutions, but it is not necessarily true that they glue together to a global solution. This point of view is useful because is more common to have an action of a ring $R$ on the space of solutions. For example, if they have some symmetry under the action of a group and stable for multiplication by a real constant, the ring $\mathbb{R}[G]$ is acting on the space of solutions; however in this situation it is much more natural to have a ""continous"" group of symmetries (like rotations) , and one considers the space of solutions as a module over the associated lie algebra.
Note that the example of "generalized functions" above is a special case of the "solutions to a constraint" case: sections of the function $x^2 \mapsto x$ are exactly the complex functions that satisfy $f(x)^2 = x$.
Second point: what does it mean to be an injective sheaf? The most "neat" consequence is that its cohomology groups are all zero (except the zero-th one which coincide with the module).
Intuitively, I think as "having zero cohomology" like being without holes of any dimension. A concrete reincarnation of this intuition is the fact that an injective sheaf is "flasque", or " flabby": this means that any local section extends to a global section (read: any local solution extends tk a global solution!).
Now we see a completely different approach, which is equivalent to be injective but not pictorial as the previous.
Recall the criterion for which is enough to check that every morphism $I \to M$ extends to a morphism $R \to M$, where $I$ is an ideal of $\mathbb{R}$. Suppose the ring is noetherian, so that the ideal $I$ is generated by a finite number of elements $f_1, \ldots, f_n$. This means we have an exact sequence
$$ 0 \to K \to R^n \to I \to 0 $$
And that a map from $I$ to $M$ is the same as choosing n elements $m_1, \ldots, m_n\in M $ satisfying the relations prescribed by $K$. Finding an extension to $R$ means to find an element $m \in M$ such that $f_j m = m_j$. Let us look at an easy example: if you take an ideal generated by one element $(f) $, for any $m_1 \in M$ you can find an element $m$ such that $fm = m_1$. In other words, you have "" $ m =m_1/f$, that is the module is divisible. For PID like $\mathbb{Z}$ this is equivalent to be injective. In general, you can always find a "common divisor" given that the elements satisfy the obvious relations they should satisfy. If for example $R= \mathbb{C}[x, y]$ and you want an element $m$ such that $xm = m_1, ym= m_2$, it's necessary that $m_1, m_2$ satisfy $ym_1 = xym = xm_2$.
In terms of sheaves, this is not much more significant. Given an ideal $I=(f_1, \ldots, f_r) $, it means that if some global sections $m_1, \ldots, m_r$ satisfies the relations analogous to the relations satisfied by $f_1, \ldots, f_r$, you can always find a global section $m$ such that $f_j m = m_j$.
