It is less confusing to first see how things work for ordered pairs. Consider the following two rules:
- $\eta$-equality for pairs:
$$\frac{u : A\times B}{u \equiv (\pi_1 u, \pi_2 u)}$$
- extensionality rule for pairs:
$$\frac{v : A\times B \quad w : A\times B \quad \pi_1 v \equiv \pi_1 w \quad \pi_2 v = \pi_2 w}{v \equiv w}$$
(Note: it is important that we consider rules, which are not to be confused with corresponding inhabitants of certain types, e.g. $\Pi (u : A \times B) \,.\, \mathrm{Id}_{A \times B}(u, (\pi_1 u, \pi_2 u))$.)
There are several ways in which we can read the above rules. The $\eta$-rule states that an element of $A \times B$ does not change if we take it apart and put it back together. It also states that every element of $A \times B$ is (equal to) a pair, namely the one we get by pairing together its components. Extensionality states that elements of $A \times B$ are equal if their projections are equal.
But these two principles are inter-derivable in the presence of congruence rules and $\beta$-rules for pairs:
$\eta$-rule implies extensionality: if $\pi_1 v = \pi_1 w$ and $\pi_2 v = \pi_2 w$ then by the $\eta$-rule $v = (\pi_1 v, \pi_2 v) = (\pi_1 w, \pi_2 w) = w$.
extensionality implies the $\eta$-rule: because $\pi_1 u = \pi_1 (\pi_1 u, \pi_2 u)$ and $\pi_2 u = \pi_2 (\pi_1 u, \pi_2 u)$, by extensionality $u = (\pi_1 u, \pi_2 u)$.
Let us repeat the exercise for function types, but first we should clear up what is meant for a function to "have a value". An ordered pair does not just "have components" -- it has a first component and a second component. Likewise, a function $f : A \to B$ does not just "have values" -- it has a value at an argument. We may extract the value at $a : A$ by applying $f$ to $a$.
Here are the $\eta$-equality and extensionality rules for functions:
- $\eta$-equality for functions:
$$\frac{f : A \to B}{f \equiv (\lambda x \colon A \,.\, f\, x)}$$
- extensionality rule functions:
$$\frac{f : A \to B \quad
g : A \to B \quad
x : A \vdash f \, x \equiv g \, x}{f \equiv g}$$
(Note: at this point it is really quite important not to confuse the above function extensionality rule with what is known as function extensionality axiom, which states that $\Pi (f, g : A \to B) \,.\, (\Pi (x : A) \,.\, \mathrm{Id}_B(f x, g x)) \to \mathrm{Id}_{A \to B}(f, g)$ is inhabited.)
The situation is essentially the same as before. The $\eta$-rule states that a function does not change if we apply it to a variable and then abstract over the variable. It also says that every element of $f : A \to B$ is a function, namely the one that maps $x$ to $f \, x$. The extensionality rule says that two elements of $A \to B$ are equal if they act the same way on arguments. Another way to say this is: if $f$ and $g$ have equal values (at an arbitrary argument $x : A$), then they are equal.
I will leave it as an exercise to show that the $\eta$-equality and extensionality rules for functions are inter-derivable.
But can we read the $\eta$-rule as stating that "$f$ is determined by its values"? Yes of course: $f$ is equal to the mapping which takes an argument $x : A$ to $f \, x$ (the value of $f$ at $x$).
I hope these considerations make it clear in what sense the $\eta$-equality rule for functions states that a function is determined by its values.
P.S.: In your question you suggest an interpretation by which "have the same values" means "have the same image". This is quite nonsensical, as it matters which arguments maps to any given value in the image. After all $\sin$ and $\cos$ have the same image $[-1,1]$ but they're hardly equal.