The choice that you make can result in a huge difference in the solution. Neither method is particularly good and both can be quite unstable.
For example, suppose that
$M=[1\;0\;0\; 0;\; 0 \; 1 \; 0 \; 0; \; 0 \; 0 \; 0 \; 0;\; 0 \; 0 \; 0 \; 0]$
and
$N=[1;\; 1; \; 1; \; 1]$.
What do you want $L$ to be in this situation? Why?
Now, suppose that you try using
$\delta = 1 \times 10^{-14}$.
Then your approximation to $M^{1/2}$ is
$M^{1/2}=[1\;0\;0\; 0;\; 0 \; 1 \; 0 \; 0; \; 0 \; 0 \; 1 \times 10^{-7} \; 0;\; 0 \; 0 \; 0 \; 1 \times 10^{-7}]$
and your approximation to $M^{-1/2}$ is
$M^{-1/2}=[1\;0\;0\; 0;\; 0 \; 1 \; 0 \; 0; \; 0 \; 0 \; 1 \times 10^{7} \; 0;\; 0 \; 0 \; 0 \; 1 \times 10^{7}]$.
In this case, the least squares solution is
$L=[1; \; 1; \; 1 \times 10^{14} \; 1 \times 10^{14}]$.
In the same example, if you use the pseudoinverse, then you'd get
$L=[1; \; 1; \; 0 \; 0]$.
Using the pseudoinverse effectively eliminated from consideration any projection of $N$ onto the null space of $M$. That might be what you want to do.
However, if $M$ is ill-conditioned, then it becomes practically impossible to distinguish small eigenvalues of $M$ from $0$ eigenvalues. As a result the computation using the pseudoinverse also becomes extremely unstable and dependent on the tolerance used in computing the pseudoinverse. This isn't good either.
Continuing the example, suppose that
$M=[1\;0\;0\; 0;\; 0 \; 1 \; 0 \; 0; \; 0 \; 0 \; 1 \times 10^{-14} \; 0;\; 0 \; 0 \; 0 \; 1 \times 10^{-14}]$.
The pseudoinverse solution is then
$L=[1; \; 1; \; 1 \times 10^{14};\; 1 \times 10^{14}]$
Now, suppose that due to a rounding tolerance in computing the pseudoinverse, you compute
$\mbox{pinv}(M)=[1\;0\;0\; 0;\; 0 \; 1 \; 0 \; 0; \; 0 \; 0 \; 0 \; 0;\; 0 \; 0 \; 0 \; 0]$
Then you'd get
$L=[1; \; 1; \; 0 \; 0]$
again.
You really need to back up a bit and explain to us why you're trying to solve this particular least squares problem in a situation where $M$ might be singular. It's likely that you need to deal with this possible ill-conditioning at a higher level by using some sort of regularization approach that is based on the particular aspects of your underlying problem.
Best Answer
No, consider the following counterexample: Take $$ M = \begin{pmatrix} 1 & 2 \\ 2 & 5 \end{pmatrix} , \quad J = \begin{pmatrix} 1 & 2\end{pmatrix},$$ then $J^\# = \begin{pmatrix} 1 \\ 0\end{pmatrix}$ and your projection is given by $I - J^\# J = \begin{pmatrix} 0 & -2\\ 0 & 1\end{pmatrix}$ and this is definitely not non-negative definite by your definition.
Edit: Can't seem to get the matrices right...