I have a silly doubt. Well, if $f,g: X \to Y$ are two functions, then we can consider the equivalence relation $\sim_{f,g}$ on $Y$ defined as the equivalence relation generated by the relation
$$ R:=\{(f(x),g(x)): \,\, x \in X\} $$
What is exactly such a relation $\sim_{f,g}$? First, I pass to the symmetric extension adding the pairs $\{(g(x),f(x)): \,\, x \in X\}$. Next, can I write
$$ y \sim_{f,g} y' \iff y=y' \, \vee \, \exists x \in X \,\, s.t. \,\, y=f(x) \, and \, y'=g(x) $$
Actually, $\sim_{f,g}$ must be the transitive closure of $R$, so it should be $$ y \sim_{f,g} y' \iff \exists n \ge 0, y_0,\dots,y_n \in X \,\, s.t. \,\, y=y_0, \, y'=y_n \, and \, y_iRy_{i+1} \, \forall i=0,\dots,n-1 $$ (by abuse, I denote by $R$ the starting relation together with the symmetric and the reflexive parts).
Which of the two formulas is correct?
Both of your formulas are incorrect. For consider $X = \{0\}$, $Y = \{0, 1\}$, $f(0) = 0$, $g(0) = 1$.
Then clearly we see that $\sim_{f, g} = Y^2$.
But according to both of your statements, we should expect that $(1, 0) \notin \sim_{f, g}$.
Edit: the question has been edited, so my answer must also be edited.
If $R$ is any relation, the smallest transitive, reflexive relation containing $R$ is always $\{(a, b) | \exists n \geq 0, \exists x_0, ..., x_n (a = x_0 \land b = x_n \land \forall i (0 \leq i < n \implies x_i R x_{i + 1})\}$. So your second answer is now correct.