Injective map on Coordinate rings implies dense image?

1) The answer to the stated question is No !

For example the inclusion $$f:\mathbb P^1(k)\to \mathbb P^2(k):|x,y]\mapsto [x,y,0]$$ has as pull-back map the identity $$f^\ast\colon k[\mathbb P^2(k)]=k\rightarrow k[\mathbb P^1(k)]=k:q\mapsto q,$$ (which is certainly injective) but the image of $f$ (consisting of the points $[x,y,0]$) is certainly not dense in $\mathbb P^2(k)$.

2) However if $W$ is affine the answer is Yes!

a) The key remark here is that a subset $S\subset W$ is not dense if and only if there exists a non-zero regular function $0\neq \phi \in k[W]$ such that $\phi(s)=0$ for all $s\in S$.
Beware that this key remark is completely false for non-affine $W$ !

b) Now, if $S=f(V)$ the above condition becomes $\phi(f(v)))=0$ for all $v\in V$ or equivalently $f^\ast(\phi )=0\in k[V]$.
But if $f^\ast$ is injective it is impossible to have both $0\neq \phi \in k[W]$ and $f^\ast(\phi )=0\in k[V]$, so that the inexorable conclusion under this injectivity hypothesis is that $f(V)$ is dense in $W$.

[By the way, irreducibilty of any variety in sight is irrelevant]

Edit: a justification for 2) a)
If $S\subset W$ is not dense, there is a closed subset $C\subsetneq W$ with $S\subset C$.
And since $C=V(\phi_1,\cdots, \phi_r)=\bigcap V(\phi_i)$ for some $\phi_i\in k[W]$, we may assume (by possibly taking a bigger $C$) that $C=V(\phi)$.
Now $\phi \neq 0$ since we have $C=V(\phi)\neq W$ and we have found the promised $\phi \in k[W]$.

Conversely if some $\phi\neq 0\in k[V]$ vanishes on $S$, then $S\subset V(\phi)\subsetneq W$ and thus $\bar S\subset V(\phi)\subsetneq W$, which implies that $S$ is not dense.

Prop: Let $F:V\rightarrow W$ be a morphism of affine algebraic varieties then pullback $F^{*}:\mathbb{C}[W]\rightarrow \mathbb{C}[V]$ is injective if and only if $F$ is dominant, that is, the image set $F(V)$ is dense in $W$.

Proof:

First, let us assume that the image of $V$ under $F$ is dense in $W$. If $g\in\mathbb{C}[W]$ is in the kernel of $F^{*}$ then by definition $F^{*}(g)=0$ meaning that $g\circ F$ is equivalent zero meaning that $g$ is zero on $F(V)$. Since $g$ is a polynomial and hence is continous this means it is zero on $W$ as it zero on $F(V)$, which is a dense set. Thus, the kernel of $F^{*}$ is trivial implying that $F^{*}$ is injective.

Alternatively if we assume $F^{*}$ is injective then to show $F$ is dominate we need to show that if $U\subset W$ is a nonempty open subset of W then $F(V)\cap U$ is nonempty. In the Zarski topology an open set $U$ is of the form $U=W\setminus X\cap W$ where $X$ is an affine algebraic variety. However, we know that every affine algebraic variety is equal to the intersection of a finite number of hypersurfaces, and thus it suffices to consider open sets of the form $U=W\setminus \mathbb{V}((f))\cap W$ where $f$ is a complex polynomial. Now since $U$ is nonempty $f$ is not equivalently zero on $W$ as if it were then $W\subset \mathbb{V}((f))$ implying $U=\emptyset$. Sine $F^{*}$ is injective this means it has trivial kernel and hence that $F^{*}(f)\not\equiv0$. By definition this means that $f\circ F\neq0$, and so there exists a $v\in V$ such that $f\circ F(v)\neq0$. However, by construction $f$ vanishes on precisely $W\setminus U$ and thus $F(v)$ must be in $U$. This means that $F(V)\cap U\neq\emptyset$ and thus the image of $V$ is dense in $W$. So we can conclude that the pullback $F^{*}:\mathbb{C}[W]\rightarrow\mathbb{C}[V]$ is injective if and only if $F$ is dominant.