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.