Let $n$ be the length of a string. We start with the (non-terminal) symbol $S$ which has length $n=1$.
Using $n - 1$ rules of form $(non-terminal) \rightarrow (non-terminal)(non-terminal)$ we can construct a string containing $n$ non-terminal symbols.
Then on each non-terminal symbol of said string of length $n$ we apply a rule of form $(non-terminal) \rightarrow (terminal)$. i.e. we apply $n$ rules.
In total we will have applied $n - 1 + n = 2n - 1$ rules.
example
Observe following grammar in Chomsky-normal form.
$
\begin{align}
S & \to AB \\
A & \to BC | AC\\
A & \to h|b\\
B & \to a \\
C & \to z \\
\end{align}
$
Consider following derivation
$
\begin{align}
\text{Current string} & & \text{rule applied} & & \text{#rules applied} & & \text{#length of string} \\
S & & \text{\\} & & 0 & & 1 \\
AB & & S \to AB & & 1 & & 2 \\
BCB & & A \to BC & & 2 & & 3 \\
\vdots & & \vdots & & \vdots & & \vdots \\
A\cdots CB & & \text{[multiple rules]} & & n-1 & & n
\end{align}
$
This last line represents a string containing only non-terminals. You can see that a string containing $n$ non-terminals is derived using $n-1$ rules. Let's continue. Applying $n$ rules of form $A \to a$ to each non-terminal in the string above gives you a string containing only terminals and thus a string from the language decided by the grammar. The length of the string has not changed (it's still $n$) but we applied an additional $n$ rules so in total we have applied $n-1 + n = 2n - 1$ rules.
While this explanation hopefully gives you an intuitive understanding, I think it would be an useful excercise to construct a formal proof using induction.