Let $X$ be a projective variety over a field $k$ of characteristic $0$ and let $f: X \to \mathbb{P}^d$ be a $k$morphism. Can we always find an embedding $i: X \hookrightarrow \mathbb{P}^N$ such that the morphism $g=f \circ i^{1}: i(X) \to \mathbb{P}^d$ is given by $g(x)=(g_0(x): \ldots : g_d(x))$ for all $x \in i(X) \subseteq \mathbb{P}^N$ where $g_0, \ldots, g_d$ are homogeneous polynomials of the same degree in $N$ variables and coefficients in $k$?
I agree, the answer is no in general. Let's take the usual bijection: rational maps from $X$ to projective space correspond to subspaces of the global sections of line bundles on $X$.
Now you're looking for a way of embedding your variety $X$ into projective space so that when we represent the map $f$ in coordinates, $f(x) = (g_1(x) : \dots: g_n(x))$, for $g_i$ homogeneous of the same degree $d$. This means that the $g_i$ are sections of the special line bundle: $O(d)_X$. So if there is a way of representing $f$ in this nice form, then the line bundle corresponding to $f$ must be of the special form $O(d)_X$ for some embedding of $X$ into projective space.
Now from the Veronese embedding of $\mathbb{P}^n$ , finding such an $O(d)$ is the same as finding an $O(1)$. The punchline being that $f$ can be represented in this form if and only if $f$ comes from a line bundle $L$ where $L = O(1)_X$ for some embedding of $X$ into projective space (i.e. $L$ is a very ample line bundle).
Now there are many examples of line bundles which are not very ample, and any map coming from those line bundles will not be able to be written in this form. For instance, if you take an elliptic curve, no matter how you embed it into projective space, a hyperplane will intersect it in more than one point. Thus, the line bundle coming from the divisor of a point on an elliptic curve gives a map that cannot be written in this form.

$\begingroup$ I can't follow your counterexample. Can you specify the map more precisely? $\endgroup$ – Alison Miller Jun 12 '14 at 17:24

$\begingroup$ Sure: My counterexample is any map to projective space that comes from a line bundle that is not very ample. A specific not $\endgroup$ – Phil Tosteson Jun 12 '14 at 17:34

1$\begingroup$ My counterexample is any map to projective space that comes from a line bundle that is not very ample. If you want a map that is regular everywhere, then my elliptic curve is probably not a great counterexample (sorry!) This link: math.stackexchange.com/questions/542751/…, says that the canonical bundle of a hyperelliptic curve with g \geq 2 does the trick. Once you have the bundle, take generators for the vector space of global sections (here they will be differential forms), and use them to make your map $\endgroup$ – Phil Tosteson Jun 12 '14 at 17:41

$\begingroup$ OK. The question asks for a morphism of varieties, so the map should be regular everywhere. (I think you do want to work with global sections, not rational sections.) $\endgroup$ – Alison Miller Jun 12 '14 at 18:15

$\begingroup$ Totally right. Thanks for pointing that out $\endgroup$ – Phil Tosteson Jun 12 '14 at 18:19
I think the answer is no. Put $L:= f^*\mathcal{O}_{\mathbb{P}^d}(1)$, and let $m=\deg(g_i)$. You want that $f$ factors as $X \stackrel{i}{\rightarrow} U\stackrel{g}{\rightarrow} \mathbb{P}^N$, where $g$ is the rational map defined by $(g_0,\ldots ,g_d)$ and $U$ its domain of definition. This implies $L\cong i^*g^*\mathcal{O}_{\mathbb{P}^d}(1)\cong i^*(\mathcal{O}_{\mathbb{P}^N}(m)_{U})$, in particular $L$ must be divisible by $m$ in $\mathrm{Pic}(X)$. So if $L$ is not divisible in $\mathrm{Pic}(X)$, the only possibility is $m=1$, which means that $f$ is obtained from $i(X)$ by a linear projection; this can be ruled out if you assume that $\ f^*:H^0(\mathbb{P}^d, \mathcal{O}_{\mathbb{P}^d}(1)) \rightarrow H^0(X,L) $ is an isomorphism.
So, for instance, a double covering of $\mathbb{P}^d$ branched along a hypersurface of degree $2e\geq 4$ gives a counterexample.