Using importance sampling to simulate the mean of a normal distribution truncated to interval [0,1] So in these notes it says that importance sampling is:
$$\int_F sf(s)ds = \int_G s \frac{f(s)}{g(s)}g(s)ds$$
And then it proceeds to give the following example:



In this example, if we draw from $f(x)$, are we effectively drawing from the truncated standard normal distribution? Also, can someone explain why it says that $g(x)=1$ if we draw $x$ from $U[0,1]$?

This is a 'toy' example because it is easier and better to do
numerical integration, roughly as used to make printed normal tables,
to get the correct answer. However, it is a nice simple example to
get you acquainted with acceptance sampling.
Numerical integration (no simulation). So at the start, let's find the correct answer to your problem with
numerical integration. In R statistical software, this can be done
as follows:
integrand = function(x){x*dnorm(x)/diff(pnorm(c(0,1)))}
# 'dnorm' is std normal PDF;  'pnorm' is std normal CDF
integrate(integrand, 0, 1)
0.4598622 with absolute error < 5.1e-15

My 'function' is $xK\varphi(x),$ where $1/K = \int_0^1 \varphi(x)\,dx$
and $\varphi$ denotes the standard normal density.
So for $X$ distributed according to your truncated normal distribution
with denisty $\varphi^*(x),$ for $x \in (0,1)$ and $0$ otherwise,
we have $E(X) = \int_0^1 x\varphi^*(s)\,dx \approx 0.4599.$ [Reality check: A sketch
should convince you that the answer must be in $(0,1)$ and slightly below $1/2.$]
Brute force simulation. In R, the 'brute force' simulation method you mention amounts to the following:
x = rnorm(10^6)          #'rnorm' samples from std normal
mean(x[x > 0 & x < 1])
## 0.4598828

This is indeed an inefficient method because we are averaging over only
fewer than 341,000 out of the one million sampled values of x. (This inefficiency  is to
be anticipated, because $1/K = P(0 < Z < 1) = .3413,$ where 
$Z \sim \mathsf{Norm}(0,1).$ ) I got a good answer because I used
a million iterations, which would have been an unthinkable extravagance
only a few years ago. 
sum(x > 0 & x < 1)
## 340532

Importance sampling.  In answer to one of your questions: 
$g(x) = 1,$ for $x \in (0,1)$ because
that is the PDF of $\textsf{Unif}(0,1).$ 
Now, consider the equation
$$\int_0^1 xf(x)\,dx = \int_0^1 x\frac{f(x)}{g(x)}g(x)\,dx = \int_0^1 xw(x)g(x)\,dx
= \int_0^1 xw(x)\,dx,$$
where $w(x) = f(x)/g(x).$ Here $f(x) = \varphi^*(x)$ above.
R code for the desired mean below uses all one million values sampled
from $\mathsf{Unif}(0,1).$
K = 1/diff(pnorm(c(0,1)))  
u = runif(10^6)
K*mean(u*dnorm(u))
## 0.4599921

