Wikipedia's page on Ito's lemma introduces random process $dX_{t}=\mu _{t}\,dt+\sigma _{t}\,dB_{t}$, where $\mu_t$ and $\sigma_t$ are deterministic, and $B_t$ is a Wiener process.
Being a non-mathematician, I latched on to the the informal derivation of Ito's lemma:
$$ df = \left({\frac {\partial f}{\partial t}}+\mu _{t}{\frac {\partial f}{\partial x}}+{\frac {\sigma _{t}^{2}}{2}}{\frac {\partial ^{2}f}{\partial x^{2}}}\right)dt+\sigma _{t}{\frac {\partial f}{\partial x}}\,dB_{t} $$
I squared this off with the section on Geometric Brownian motion (GBM), where
$$ dS_{t}=\sigma S_{t}\,dB_{t}+\mu S_{t}\,dt $$
and $f(S_t)=\log(S_t)$. It starts with
$$ df = f'(S_t) dS_t + \frac{1}{2} f''(S_t)(dS_t)^2 $$
and ends up with
$$ df = \sigma \,dB_{t}+\left(\mu -{\tfrac {\sigma ^{2}}{2}}\right)\,dt $$
If I slavishly start with Ito's lemma above and calculate $df$ by precalculating $\partial f/\partial t$, $\partial f/\partial x$, and $\partial^2 f / \partial x^2$, I get the same result. However, I have to make use of the fact that $\sigma_t\triangleq\sigma S_t$ and $\mu_t\triangleq\mu S_t$. Since $S_t$ is stochastic, so are $\sigma_t$ and $\mu_t$. This violates the condition stated above that they are deterministic.
Can someone please explain why this is consistent? I come from a background of engineering rather than math, so something intuitive would be nice.
Deterministic is too strong of a requirement. You only need $t \mapsto \sigma_t$ and $t \mapsto \mu_t$ to be adapted to the Brownian motion. Intuitively, what this means is that the processes are not able to "peek into future" values of $B$.