It is in fact true that the normalization of a projective variety is projective, as J.C. Ottem discusses in the comments.

It is not true that if a normal variety is mapped to a projective space by a linear series $V\subset H^0(L)$ then some larger linear series $W\supset V$ has image isomorphic to the normalization.

For instance, let $C$ be a general smooth curve of genus $g \gg 0$, and pick a general line bundle $L$ of degree $g+2$. By Riemann-Roch, $h^0(L) = 3$, and thus the map induced by $|L|$ maps $C$ to $\mathbb{P}^2$. For large enough $g$, however, the general curve of genus $g$ is not isomorphic to a smooth plane curve, and thus the image cannot be smooth. Moreover, we're using the complete series of sections of $|L|$ already, so there aren't "more sections" to include.

However, the following modification *is* true. Suppose $X\subset \mathbb{P}^n$ is a variety, with normalization $f:\overline{X}\to X$. Then there *exists* a line bundle $L$ on $X$ such that $f$ is given by a collection of sections of $L$ and the complete series $|L|$ gives an embedding of $\overline{X}$ into some big projective space.

Roughly, if $L = f^* \mathcal{O}(1)$, we can modify $L$ by adding a sufficiently ample divisor $nH$ so that $L+nH$ gives an embedding. But if $V \subset H^0(L)$ corresponds to $f$, then multiplying by a fixed section $D$ of $nH$ gives us an inclusion $D + V \subset H^0 (L+nH)$; note that this series has $D$ as a base locus. The map corresponding to this series is just $f$, realized as a projection from the big projective space which $L+nH$ maps $\overline{X}$ to.