Because its Lyapunov exponent is zero, but that is not what you are computing.

Instead of looking at random walks on $SL(2, \mathbb{R})$, let me focus on random walks on $\mathbb{R}_+^*$, as there is the same issue. Let $(X_n)$ be i.i.d. in $\mathbb{R}_+^*$, and to make things simple, assume that there are only finitely many values. Let $P_n := X_n \ldots X_1$.

The Lyapunov exponent of this random walk is the real $\Lambda$ such that

$$\lim_{n \to + \infty} \frac{\ln (P_n)}{n} = \Lambda.$$

By the law of large numbers, $\Lambda = \mathbb{E} (\ln(X_1))$. For instance, if $X_1$ takes values $2$ and $1/2$ each with probability $1/2$, the Lyapunov exponent is $0$: the markov chain $(P_n)$ will oscilalte between very large and very low values.

However, if you compute the expectation of the norm, a short computation gets you $\mathbb{E} (P_n) = (5/4)^n$, which grows exponentially fast. But that does not mean that the Lyapunov exponent is $\ln (5/4)$. The issue is merely that the exponential does not commute with the expectation:

$$1 = e^{\mathbb{E}(\ln(P_n))} \leq \mathbb{E} (e^{\ln(P_n)}) = (5/4)^n.$$

To get back to a general random walk, we have $\ln(P_n) \simeq \mathcal{N} (n\mu, n\sigma^2)$. The Lyapunov exponent is the constant $\mu$. However,

$$\mathbb{E} (P_n) \simeq \mathbb{E} (e^{\mathcal{N} (n\mu, n\sigma^2)}) = \mathbb{E} (e^{n(\mu+\frac{\sigma^2}{2})}),$$

so taking the norm as you get gives you an error of $\sigma^2/2$ coming only from the diffusion of the random walk.

Because its Lyapunov exponent is zero, but that is not what you are computing.

Instead of looking at random walks on $SL(2, \mathbb{R})$, let me focus on random walks on $\mathbb{R}_+^*$, as there is the same issue. Let $(X_n)$ be i.i.d. in $\mathbb{R}_+^*$, and to make things simple, assume that there are only finitely many values. Let $P_n := X_n \ldots X_1$.

The Lyapunov exponent of this random walk is the real $\Lambda$ such that

$$\lim_{n \to + \infty} \frac{\ln (P_n)}{n} = \Lambda.$$

By the law of large numbers, $\Lambda = \mathbb{E} (\ln(X_1))$. For instance, if $X_1$ takes values $2$ and $1/2$ each with probability $1/2$, the Lyapunov exponent is $0$: the Markov chain $(P_n)$ will oscillate between very large and very low values.

However, if you compute the expectation of the norm, a short computation gets you $\mathbb{E} (P_n) = (5/4)^n$, which grows exponentially fast. But that does not mean that the Lyapunov exponent is $\ln (5/4)$. The issue is merely that the exponential does not commute with the expectation:

$$1 = e^{\mathbb{E}(\ln(P_n))} \leq \mathbb{E} (e^{\ln(P_n)}) = (5/4)^n.$$

To get back to a general random walk, and very roughyl, we have $\ln(P_n) \simeq \mathcal{N} (n\mu, n\sigma^2)$. The Lyapunov exponent is the constant $\mu$. However,

$$\mathbb{E} (P_n) \simeq \mathbb{E} (e^{\mathcal{N} (n\mu, n\sigma^2)}) = \mathbb{E} (e^{n(\mu+\frac{\sigma^2}{2})}),$$

so taking the norm as you get gives you an error coming from the diffusion of the random walk (well, in practice, the exact value of $\sigma^2/2$ for this error is wrong, but I don't think the heuristics is too bad at this level).