To truncate a distribution is to restrict its values to an interval and re-normalize the density so that the integral over that range is 1.
So, to truncate the $N(\mu, \sigma^{2})$ distribution to an interval $(a,b)$ would be to generate a random variable that has density
$$ p_{a,b}(x) = \frac{ \phi_{\mu, \sigma^{2}}(x) }{ \int_{a}^{b} \phi_{\mu, \sigma^{2}}(y) dy } \cdot \mathcal{I} \{ x \in (a,b) \} $$
where $\phi_{\mu, \sigma^{2}}(x)$ is the $N(\mu, \sigma^2)$ density. You could sample from this density in a number of ways. One way (the simplest way I can think of) to do this would be to generate $N(\mu, \sigma^2)$ values and throw out the ones that fall outside of the $(a,b)$ interval, as you mentioned. So, yes, those two bullets you listed would accomplish the same goal. Also, you are right that the empirical density (or histogram) of variables from this distribution would not extend to $\pm \infty$. It would be restricted to $(a,b)$, of course.
In the univariate case, when we truncate a zero-mean normal at zero (which is the same as taking its absolute value), then the folded-normal and the truncated normal family share a common member, the half-normal.
This is not in general the case in a multivariate setting. Treating the bivariate case, let's see under which conditions the identity nevertheless holds.
The density of the bivariate half normal can be compactly written (adjusting the formulas that can be found here ) as
$$f_H(x,y)=4\frac{\sqrt {1-\rho^2}}{s_xs_y}\cdot\phi(x/s_x)\cdot\phi(y/s_y)\cdot\text{cosh}\left(\frac{\rho xy}{s_xs_y}\right)$$
with
$$s_x = \sigma_x\sqrt {1-\rho^2}, \,s_y = \sigma_y\sqrt {1-\rho^2}$$
and where $\phi()$ is the standard normal PDF and $\text{cosh}$ is the hyperbolic cosine function.
The density of the bivariate normal truncated from below at zero is
$$f_T(x,y) = \left[\int_{0}^{\infty}\frac {1}{\sigma_x}\phi(x/\sigma_x)\Phi\left(\frac {\rho}{\sqrt {1-\rho^2}}\frac {x}{\sigma_x}\right)dx\right]^{-1}\cdot\frac{\sqrt {1-\rho^2}}{s_xs_y}\times\\\times\; \phi(x/s_x)\cdot\phi(y/s_y)\cdot \exp\left\{-\frac{\rho xy}{s_xs_y}\right\}$$
Note that multiplying and dividing the integral in the expression by $2$ we obtain
$$\Big[...\Big] = \frac 12-\frac 12 G\left(0;0,\sigma_x,\frac {\rho}{\sqrt {1-\rho^2}}\right)$$
where $G(x;\text{location},\text{scale}, \text{skew})$ is the Skew Normal CDF, for which we have (see for example here eq. $(18)$)
$$G\left(0;0,\sigma_x,\frac {\rho}{\sqrt {1-\rho^2}}\right) = 2B(0;0;-\rho)$$
where $B()$ is the bivariate standard normal integral. This relation verifies that the integral can equivalently be written in terms of the $y$ variable without imposing equality of variances between $x$ and $y$ -the variance term does not affect its value.
Then, if $\rho =0$, one can verify that the two densities become identical: the reciprocal of the integral equals $4$ and the other generally non-equal terms all become unity, giving
$$\rho =0 \Rightarrow f_H(x,y) = f_T(x,y) = \frac{4}{\sigma_x\sigma_y}\cdot\phi(x/\sigma_x)\cdot\phi(y/\sigma_y)$$
So it appears that the condition needed for the identity of the densities to hold is that the two underlying zero-mean normals are independent, but not in addition that they have the same variance.
Opportunity taken, Horrace 2005 collects many useful results related to multivariate truncated normal distributions. Perhaps the most interesting is the following:
If $\mathbf w$ is a multivariate truncated normal, then the linear
combination $\mathbf y = \mathbf D\mathbf w+\mathbf b$ is not
multivariate truncated normal, excpet if $\mathbf D = \mathbf I$.
In other words, with truncation present, normality is preserved under shifting but not under scaling.
Also,
If $\mathbf w$ is a multivariate truncated normal vector of correlated
random variables, then the marginal distributions are not, in general,
truncated normal, but the conditional distributions are truncated
normal. If the variables are independent, then the marginal distributions also are truncated normals.
Finally, log-concavity and the multivariate generalization of unimodality characterize the multivariate truncated normal.
Best Answer
Yes, the approaches give the same results for a zero-mean Normal distribution.
It suffices to check that probabilities agree on intervals, because these generate the sigma algebra of all (Lebesgue) measurable sets. Let $\Phi$ be the standard Normal density: $\Phi((a,b])$ gives the probability that a standard Normal variate lies in the interval $(a,b]$. Then, for $0 \le a \le b$, the truncated probability is
$$\Phi_{\text{truncated}}((a,b]) = \Phi((a,b]) / \Phi([0, \infty]) = 2\Phi((a,b])$$
(because $\Phi([0, \infty]) = 1/2$) and the folded probability is
$$\Phi_{\text{folded}}((a,b]) = \Phi((a,b]) + \Phi([-b,-a)) = 2\Phi((a,b])$$
due to the symmetry of $\Phi$ about $0$.
This analysis holds for any distribution that is symmetric about $0$ and has zero probability of being $0$. If the mean is nonzero, however, the distribution is not symmetric and the two approaches do not give the same result, as the same calculations show.
This graph shows the probability density functions for a Normal(1,1) distribution (yellow), a folded Normal(1,1) distribution (red), and a truncated Normal(1,1) distribution (blue). Note how the folded distribution does not share the characteristic bell-curve shape with the other two. The blue curve (truncated distribution) is the positive part of the yellow curve, scaled up to have unit area, whereas the red curve (folded distribution) is the sum of the positive part of the yellow curve and its negative tail (as reflected around the y-axis).