I asked this question What is the importance of $R$ being a field in this question? here and I was told by Connor that if $R$ is not a field then the LES axiom will not be satisfied but I do not understand why and how. Could anyone show me the details of this, please?
Also, I am wondering how will the suspension axiom proved in our case in the problem? I.e. when $R$ is a field. Could anyone help in this, please?
Here is the statement of the problem in the mentioned question above:
Problem 22.39. Suppose $R$ is a field.
(a) Show that $h^n(?) = \operatorname{Hom}_R( H_n(? ; R), R)$ is a cohomology theory defined on (at least) the category of finite CW complexes.
(b) Show that $u$ is a natural transformation of cohomology theories.
(c) Prove Theorem 22.37.
Since singular homology $H_n$ satisfies the Eilenberg-Steenrod axioms for homology, we know that an inclusion $A\hookrightarrow X$ induces an LES in homology. Now, you want to use this LES and plug it into your $h^n$ and get another LES. The problem is that this is equivalent to applying Hom to an exact sequence and getting an exact sequence out. However, Hom is only left exact in general.
As Connor commented on your other question, Hom is also right exact if and only if the object you're taking Hom $\textit{into}$ is injective (i.e., $\text{Hom}(-,A)$ is right exact if and only if $A$ is injective). You can check that every field is injective as a module over itself, so $R$ being a field is sufficient to conclude that Hom of the LES in singular homology is also an LES.
edit: I realize I didn't address the suspension part of your question. I don't believe the existence of suspension isomorphisms is an axiom for ordinary homology theories. It should follow from an application of the Mayer-Vietoris sequence, which itself follows from the Eilenberg-Steenrod axioms. In any case, by the argument above, we know that if $R$ is a field then $\text{Hom}(-,R)$ will be exact. Then since singular homology $H_n$ has suspension isomorphisms, we can apply $h^n$ to the corresponding short exact sequences and obtain suspension isomorphisms for $h^n$ as well.
second edit: In response to the comment about showing how the suspension isomorphisms can be obtained via the Mayer-Vietoris sequence, let $X = \Sigma Y$ be the suspension of $Y$. Note that $X$ can also be written as $C(Y)\cup C(Y)/\sim$, where $\sim$ simply identifies the nontrivial end of the cone $C(Y)$ with the nontrivial end of the other $C(Y)$ (i.e., if $Y = S^1$, the $C(Y)$ is just a hollow cone with hollow circular bottom, and this circular bottom is the "nontrivial" end I'm describing above). Then letting $A\subset X$ consist of one of the cones $C(Y)$ along with "a little extra" past the equator and $B\subset X$ be similarly defined with the other cone $C(Y)$, we see that the interiors of $A$ and $B$ cover $X$. Furthermore, $A\cap B$ will be a cylinder homeomorphic to $[0,1]\times Y$, meaning $A\cap B$ is homotopy equivalent to $Y$. Plugging all of this into the Mayer-Vietoris sequence gives the desired isomorphisms when you use the fact that $A$ and $B$ are both contractible.