It's perhaps best to think about this first in the context of a particular structure and a particular formula, where you can easily see what's going on. Consider the structure of the natural numbers with the usual ordering relation $<$. The sentence $\forall x\exists y\,(x<y)$ is true in this structure, because for every natural number $x$ there is a larger one $y$, for example $x+1$ (or $x+17$ or $(2x)!$ or lots of other choices). Equivalently, the second-order sentence $\exists F\forall x\,(x<F(x))$ is true in the same structure, witnessed by functions like $F(x)=x+1$ (or $F(x)=x+17$ or $F(x)=(2x)!$, or lots of other functions $F$).
On the other hand, we cannot say (as you did) that $\forall x\exists y\,(x<y)\models\forall x\,(x<F(x))$. To see this, notice first that this "logical consequence" relationship concerns structures for a language that has not only the relation symbol $<$ but also the function symbol $F$. Consider the structure of natural numbers with $<$ as before and some interpretation of the function symbol $F$. Then $\forall x\exists y\,(x<y)$ is true (regardless of the interpretation of $F$ because it doesn't mention $F$), but whether $\forall x\,(x<F(x))$ is true depends on how we interpret $F$. It's true if we take $F(x)=x+1$ (or $F(x)=x+17$ or $F(x)=(2x)!$, or lots of other functions $F$), but it's false if we take $F(x)=0$ (or $F(x)=17$ or $F(x)=\lfloor x/2\rfloor$, or lots of other functions). But $\models$ means that every structure satisfying what's on the left of $\models$ also satisfies what's on the right. So in our case, it would require that $\forall x\,(x<F(x))$ is true regardless of how we interpret $F$ as long as $\forall x\exists y\,(x<y)$ is true. Since that doesn't hold when $F$ is "badly" interpreted ($0$ or $17$ or $\lfloor x/2\rfloor$), we don't have $\forall x\exists y\,(x<y)\models\forall x\,(x<F(x))$. The most we can say is that, having satisfied $\forall x\exists y\,(x<y)$, we can, by suitably interpreting $F$, also satisfy $\forall x\,(x<F(x))$ --- and vice versa. That's what "equisatisfiable" means.
Basically, the distinction is between talking about a specific situation versus all possible situations.
Suppose I have two different propositional variables $p$ and $q$. Then:
$p$ and $p\wedge p$ are logically equivalent. (Remember that "$\wedge$" means "and.")
$p$ and $q$ are not logically equivalent.
However, $p\iff q$ might be true (e.g. if both $p$ and $q$ happen to be true).
This is ultimately a distinction between talking about general necessities versus specific situations. The keyword here is "model." In the setting of propositional logic (there are other logics), a model is just a specific assignment of truth values to the propositional variables in the language. E.g. suppose our language has propositional atoms $p, q, r$. Then "$p$ and $q$ are true, $r$ is false" (or rather, the function $\nu: \{p, q, r\}\rightarrow\{true, false\}$ sending $p$ and $q$ to $true$ and $r$ to $false$) is a model. Note that given a model, we can also talk about the truth values of more complicated sentences in that model: e.g. "$p\wedge r$" is false according to the model above.
(Indeed, we can prove by "structural induction" that an assignment of truth values to propositional variables uniquely extends to an assignment of truth values to all propositions, which respects the obvious rules - e.g. if $\varphi$ and $\psi$ are both assigned "true," then $\varphi\wedge\psi$ must be assigned "true," and so forth. Sometimes a model is defined as a truth assignment to all propositions, which satisfies these reasonable rules; the proof described in the previous sentence means that we can get away with the simpler definition above.)
When we say that two sentences are logically equivalent, we mean that there is no model in which they have different truth values. The expression "$p\iff q$," however, is a (compound) sentence which is (in general) true in some models and false in others. This is the distinction:
The notion of "logical equivalence" is talking about what things are possible in general.
When we say that a sentence is true/false, we are talking about its truth/falsity in a specific model.
For example, in the model $\nu$ defined above the sentence "$p\iff q$" is true, even though $p$ and $q$ are not logically equivalent (exercise).
At this point, it's useful to introduce a bit of terminology: a sentence which is true in every model is called a tautology. When we say "$\varphi$ and $\psi$ are logically equivalent," we're just saying "$\varphi\iff \psi$ is a tautology."
A bit of more advanced material
There is also a "relative" version of this. Suppose $\varphi$ is some proposition which is true in every model in which the proposition $\psi$ is also true. (Note that this just means that the proposition "$\psi\implies\varphi$" is a tautology.) Then we write "$\psi\models\varphi$."
The value of this new symbol is that it lets us generalize considerably: if $\Gamma$ is a set of propositions, we write "$\Gamma\models\varphi$" iff $\varphi$ is true in every model where every proposition in $\Gamma$ is true. If $\Gamma$ is infinite, this is meaningfully different from just talking about tautologies (since "$\Gamma\implies\varphi$" isn't actually a proposition).
However, one of the most important theorems in logic - the compactness theorem - states that if $\Gamma\models\varphi$ then there is some finite subset $\{\gamma_1, \gamma_2,...\gamma_n\}\subseteq\Gamma$ such that $\{\gamma_1, \gamma_2,...,\gamma_n\}\models\varphi$. And this just means that the proposition "$(\gamma_0\wedge\gamma_1\wedge...\wedge\gamma_n)\implies\varphi$" is a tautology. So via the compactness theorem we can reduce questions about the relation "$\models$" to questions about tautologies, but that's far from obvious.
(And there are important logics which don't have this property, so actually they are meaningfully different in general.)
Best Answer
In oder to "manufacture" a counterexample for two formuale being equisatisfiable but not equivalent, consider the first-order language of arithmetic with the predicate $<$ ("less-than") and a new (unary) function symbol $f$, and let :
$\phi := \forall x \exists y (x < y), \psi := \forall x (x < f(x))$.
The two are satisfiable in the same model, considering $\mathbb N$ and interpreting $<^N = <$ and $f^N: \mathbb{N} \to \mathbb{N}$ being the addition $x \mapsto x+1$.
However, they are not equivalent, i.e. there is another model where one is true and the other is not: consider again $\mathbb N$ and interpret the function symbol $f$ with : $f^N : \mathbb N \to \mathbb N$ such that $f^N(n)=0$, for all $n \in \mathbb N$. In this model, $x < f(x)$ is false.
In fact: $\forall x (x < f(x)) \vDash \forall x \exists y (x < y)$ but $\forall x \exists y (x < y) \nvDash \forall x (x < f(x))$.