I am trying to prove the following:
Let $X$ be a (finite type, pure dimension $d$) $k-$scheme for $k$ a field, and $l$ a field extension of $k$, with $X_l := X\times_k l$. Then $X$ is smooth over $k$ if $X_l$ is smooth over $l$.
Here smooth means that the sheaf of Kahler differentials is locally free.
Working locally on $X$, we take an affine piece given by $\operatorname{Spec}(B)$ for $B = k[X_i]/(f_j)$, so that the corresponding piece of $X_l$ is given by $B_l =l[X_i]/(f_j)$. Then the differentials are given by the cokernels of the linear maps corresponding to the Jacobian matrix for the $f_j$, considered as maps of $B$ and $B_l$ modules respectively. Since $B$ isn't a field, it's not clear to my why freeness of the cokernel of $Jac(f) : B_l^m\rightarrow B_l^n$ implies the freeness of $Jac(f) : B^m \rightarrow B^n$, although the converse is clear. I'm pretty sure we only need that $l/k$ is a flat extension of rings, but I don't know how to proceed.
It may even be that something much simpler to state is true:
Let $B$ be a finitely generated $k$ algebra, $l$ a field extension of $k$ and $M$ a $B-$ module. Then $M$ is free (of finite rank) iff $M\otimes_k l$ is free (of finite rank) as a $B_l$ module.
But I'm not sure.
You have more or less said it. Any field extension is not just flat, but faithfully flat. If $A\to B$ is faithfully flat and $M$ is a finitely generated $A$-module, then $M$ is projective over $A$ if and only if $M\otimes_A B$ is projective over $B$.
The last question is not quite correct. Only projectivity is preserved, not freeness (which is enough for your smoothness arguments). For example, the tangent bundle of the real sphere is projective, not free, but when you extend the field to complex numbers, it is free.