Why it is difficult to define cohomology groups in Arakelov theory?

The special fiber of $X$ might be a union of one or more irreducible curves. The local ring at the generic point of each of those points is a discrete valuation ring (being a regular local ring of dimension $1$), and so it defines a valuation on its field of fractions, which is $K(X)$.

Let's consider a special case where the difficulty already appears. Takes $V=\mathbb Z_p, K= \mathbb Q_p, X = \mathbb P^1_{\mathbb Z_p}, K(X) = \mathbb Q_p(x)$. There is a natural way to extend the valuation on $\mathbb Q_p$ to a valuation of the field $K(X)$. To do this we write an element in $\mathbb Q_p(x)$ as a ratio of two polynomials in $\mathbb Z_p[x]$, and we define the valuation to be the highest power of $p$ dividing the numerator minus the highest power of $p$ dividing the denominator. Note that, when we restrict this valuation to $\mathbb Q_p$, we recover the standard valuation on $\mathbb Q_p$.

Furthermore, to explain why this is so natural, note that for any $x$ in $\overline{\mathbb F}_p$ where neither the numerator or the denominator vanishes (mod $p$) on $x$, for any lift of $x$ to $\overline{\mathbb Z}_p$, the valuation of the rational function calculated this way is the same as the valuation when we evaluate it at $x$, simply because the numerator and denominator remain $p$-adic units in this case. Because the nonvanishing of the numerator and the denominator is a generic condition, we say this is the valuation associated to the generic point.

The theorem that rules out an analogous construction at the infinite place is the Gelfand-Tornheim theorem, which says that there is no absolute value on $\mathbb C(X)$ for $X$ a curve over $\mathbb C$ that extends the absolute value on $\mathbb C$.