For converting english statements to logical statements, I've understood that $\forall$ uses $\rightarrow$ and $\exists$ uses $\land$ (generally). I referred a particular question on this site, but I lost track of it (apologies).
Considering the same example used in the reference, "some real numbers are rational",
What is the difference between
$\exists x ( f(x) \rightarrow g(x) )$ and $\exists x ( f(x) \land g(x) )$ ?
where, $f(x) = $"$x$ is real" and $g(x) = $"$x$ is rational"
Also, where can $\exists$ be used with $\rightarrow$ and $\land$ ?
I understood that I could depict $f(x) \rightarrow g(x)$ as $\neg f(x) \vee g(x)$ and how this could prove why a conditional would work with $\forall$. But I could not figure out how and where could $\rightarrow$ and $\land$ be used with $\land$.
In logic, the symbols $\forall$ (for all) and $\exists$ (exists) are called quantifiers, whereas the symbols $\wedge$ and $\rightarrow$ (among others) are called connectors. Each quantifier may be used with any connector, and using one instead of the other results in a different proposition. Thus, I do not agree with your first claim:
Regarding the examples you ask about. Let $f(x)$ be the proposition "$x$ is real" and let $g(x)$ be the proposition "$x$ is rational."
$$\exists x: (f(x) \rightarrow g(x)) \qquad \textrm{(A)}$$
This may be read as "there is at least one $x$ such that, if $x$ is real, then $x$ is rational." This proposition is true, take for example $x=1$: $f(1)$ is true, $g(1)$ is true, then $f(1) \rightarrow g(1)$ is true. The truth of this implication derives from the definition of the logical connector $\rightarrow$ -you may check it with a truth table- and has little to do with mathematical properties of real and rational numbers.
The other proposition:
$$\exists x: (f(x) \wedge g(x)) \qquad \textrm{(B)}$$
is also true and you make take again $x=1$ to prove the existence. The truth table is different from the previous one, but yields the same truth value as before (in this case.)
These two cases may not be helpful to get the difference between both connectors. Let's build a different example. It can be easily shown that for any $f$ and $g$, if proposition (B) is true, then (A) is also true. So you could only get different truth values by getting (B) false and (A) true, for which is necessary and sufficient that $f(x)$ is false for all $x$.
Such an example may be achieved by considering $f(x) = $ " $x$ is a real solution to the equation $x^2 = -1$" and $g(x)$ may be whatever you want, e.g. $g(x) = $ "$1 \neq 0$" of $g(x) = $ "$1 = 0$", or $g(x) = $ "Thick as a Brick is the best album ever." In any of these cases (A) is true and (B) is false.