Immediate extension of valuation on an algebraically closed field is determined by cofinal sequence. Let $k$ be an algebraically closed field, let $v: k \twoheadrightarrow \Gamma \cup \{\infty\}$ a valuation on that field for some ordered abelian group $\Gamma$. Now let $w: k(X) \twoheadrightarrow \Gamma' \cup \{\infty\}$ be an extension for an ordered abelian group $\Gamma'$ such that the residue fields $kv$ and $kw$ are equal and $\Gamma=\Gamma'$ (i.e. the extension is immediate). Equivalently the set $\{w(X-c) \mid c \in k \}$ has no maximal element.
Now let $(c_{\alpha})_{\alpha < \lambda}$ be some sequence for an limit ordinal $\lambda$, such that $\gamma_{\alpha}=w(X-c_{\alpha})$ and such that $(\gamma_{\alpha})_{\alpha < \lambda}$ is cofinal in $\{w(X-c) \mid c \in k \}$
Now to my question: Why is $w$ uniquely determined by $w(X-c_{\alpha})=\gamma_{\alpha}$? I know that I somehow have to proof that $w(X-c)$ is determined for every $c \in k$ that would conclude the proof since $k$ is algebraically closed, but I could not really get there. This was marked as an easy exercise in the lecture notes I am currently reading and it certainly is, but I am a bit stuck right now. (Corollary 4.4. of: https://www.uni-muenster.de/imperia/md/content/logik/hils/mhnotes_v3.pdf)
 A: Let me try to explain the intuition first. It can be useful to imagine a valued field $(K,v)$ as an "metric space" according to the "metric" $d(a,b) = 2^{-v(a-b)}$. Of course, this makes no sense when $\Gamma$ is not a subgroup of $(\mathbb{R},+)$, but it captures the idea that two elements are "close together" when the valuation of their difference is large. 
In any valued field $(K,v)$, we have $$v(a-b) \geq \min(v(a),v(b)),$$ with equality when $v(a)\neq v(b)$. In the "metric" perspective, this turns into the ultrametric triangle inequality $$d(a,c)\leq \max(d(a,b),d(b,c)),$$ with equality when $d(a,b)\neq d(b,c)$. All "triangles" are "isosceles" with the two longest sides having equal length. 
So your sequence $c_\alpha$ with $\gamma_\alpha = w(X-c_\alpha)$ cofinal in $\{w(X-c)\mid c\in k\}$ is like a sequence converging to $X$ (or rather, getting as close to $X$ as any elements of $k$ can). And we can analyze the distance from an arbitrary element $c$ to $X$ by looking at the sequence of distances from $c$ to $c_\alpha$. In particular, once $c_\alpha$ is closer to $X$ than $c$ is, the ultrametric triangle inequality ensures that the distance from $c$ to $X$ is the same as the distance from $c$ to $c_\alpha$.

Ok, now here's the formal answer. As you noted in the question, to determine $w$ on $k(X)$, it suffices to determine $w(X-c)$ for all $c\in k$. Indeed, since $k$ is algebraically closed, every polynomial $p(X)\in k[X]$ splits as $a\prod_{i=1}^d (X-c_i)$, and $w(p(X)) = w(a) + \sum_{i=1}^d w(X-c_i)$. And every rational function $r(X)$ in $k(X)$ is a quotient $p(X)/q(X)$ of polynomials, so $w(r(X)) = w(p(X)) - w(q(X))$.  
Now fix $c$, and (by the cofinality assumption) pick $\alpha$ large enough so that $w(X-c_\alpha) > w(X-c)$. Then compute $$w(c_\alpha-c) = w((X-c)-(X-c_\alpha)) = \min(w(X-c),w(X-c_\alpha)) = w(X-c).$$
