I figured I would make an answer out the comments to the first answer addressing a point which confused me when I first learned this stuff. A morphism $f:X\to Y$ (I have to write it in this direction or else I'll confuse myself) is said to be a closed immersion if $f$ induces a homeomorphism of $X$ onto a closed subset of $Y$, and $f^\sharp:\mathscr{O}_Y\to f_*(\mathscr{O}_X)$ is surjective.
In some references I've seen it is casually remarked that the second condition is equivalent to surjectivity of the map $f_x^\sharp:\mathscr{O}_{Y,f(x)}\to\mathscr{O}_{X,x}$ for all $x\in X$. But is this really trivial? No! This map, which might reasonably be called the stalk of the morphism $f$ at $x$, is not literally the same as the stalk of $f^\sharp$ at $f(x)$. Indeed, that is a map $f_{f(x)}^\sharp:\mathscr{O}_{Y,f(x)}\to(f_*\mathscr{O}_X)_{f(x)}$. In general, there is always a natural map $\varphi_x:(f_*\mathscr{O}_X)_{f(x)}\to\mathscr{O}_{X,x}$, but it isn't in general an isomorphism. The map $f_x^\sharp$ is equal to $\varphi_x\circ f_{f(x)}^\sharp$. So while it is standard that the map $f^\sharp$ of sheaves (on $Y$!) is surjective if and only if $f_y^\sharp:\mathscr{O}_{Y,y}\to (f_*\mathscr{O}_X)_y$ is surjective for all $y\in Y$, this does not obviously say anything about surjectivity of the maps $f_x^\sharp:\mathscr{O}_{Y,f(x)}\to\mathscr{O}_{X,x}$ for $x\in X$. If however $f$ is a homeomorphism onto a closed subset $f(X)\subseteq Y$, then the stalks of $f_*\mathscr{O}_X$ at points of $Y$ are easy to compute: they are zero at points outside of $f(X)$, and at a point $f(x)\in f(X)$, we have that the natural map $(f_*\mathscr{O}_X)_{f(x)}\to\mathscr{O}_{X,x}$ is an isomorphism. So in that case, surjectivity of each $f_x^\sharp$, $x\in X$, actually will imply surjectivity of $f_y^\sharp$, $y\in Y$, and hence of $f^\sharp$.
Without the condition that $f$ is a closed topological immersion on the underlying topological spaces, it is not going to be true that $f^\sharp$ is surjective if and only if $f_x^\sharp$ is surjective for all $x\in X$. To make this clearer, let's assume $X=\mathrm{Spec}(B)$ and $Y=\mathrm{Spec}(A)$, so $f=\mathrm{Spec}(\alpha)$ for $\alpha:A\to B$ a ring homomorphism. The stalk map of $f$ at $x=\mathfrak{q}\in\mathrm{Spec}(B)$ is the ring map $A_\mathfrak{p}\to B_\mathfrak{q}$, where $\mathfrak{q}=\alpha^{-1}(\mathfrak{p})$. In general, surjectivity of this map for all $\mathfrak{q}\in\mathrm{Spec}(B)$ does not imply surjectivity of $\alpha$ itself.
I think the simplest example that will illustrate this is when $B=A_g$ is a principal localization of $A$. Then in fact the stalk map in the previous paragraph is an isomorphism for every prime ideal of $A_g$ (the set of which are in natural bijection with the set of primes of $A$ not containing $g$, i.e. $D(g)$). But the localization map $A\to A_g$ (i.e. the map on global sections of $f$) is not usually surjective. Note that in this case $f$ is a homeomorphism onto the open subset $D(g)$ of $\mathrm{Spec}(A)$, but $D(g)$ is not generally closed in $A$.
I think maybe this illustrates why the first condition is important, and why, if one wants to think about surjectivity of $f^\sharp$ in terms of the stalks of $f$, $f_x^\sharp$, for $x\in X$, the topological condition is needed, and logically "precedes" the condition on $f^\sharp$.
Lastly, I should note that the maps which I have been calling the ``stalks of $f$," $f_x^\sharp$, for $x\in X$, are in fact the stalks of the map of sheaves on $X$ (in the usual sense) $f^\flat:f^{-1}\mathscr{O}_Y\to \mathscr{O}_X$ corresponding to $f^\sharp$ under the adjunction between $f^{-1}$ and $f_*$. So surjectivity of all $f_x^\sharp$, $x\in X$, is logically equivalent to surjectivity of $f^\flat$. There is no reason to believe that $f^\flat$ is surjective if and only if $f^\sharp$ is, or even that there is an implication in either direction in general.
Example b) is not even flat, so it is not etale.
Morally, what's happening at $T=0$ in example c) is that the fiber is changing from $d$ distinct points to one point: when $d>1$, this doesn't fit with the idea of etale maps as being like covering maps. (The technical term for this is that the map ramifies at $T=0$ - as etale is equivalent to flat + unramified, this provides one proof that the map in c) is not etale.)
Before performing the computations to show that condition 3) fails for example c), I should note that the condition as written in your post at the time of this answer is not correct. It should instead read as follows:
If $f:X\to Y$ is a morphism of schemes with $y=f(x)$ for some $x\in X, y\in Y$, then $\mathfrak{m}_y\mathcal{O}_{X,x} = \mathfrak{m}_x$ and the field extension $\kappa(y)\subset\kappa(x)$ is separable.
To check flatness, we observe that $\Bbb C[T,Y]/(Y^d-T)$ is a finitely-generated free $\Bbb C[T]$ module with basis $1,Y,Y^2,\cdots,Y^{d-1}$. By computing fiber products, we see that the fiber over $T=\lambda$ is just the spectrum of the ring $R=\Bbb C[T,Y]/(Y^d-T,T-\lambda)$. But $R\cong\Bbb C[Y]/(Y^d-\lambda)$, and when $\lambda\neq0$, $Y^d-\lambda$ splits as a product of linear terms, so $R\cong \Bbb C^d$ and it is immediate to check condition 3). When $\lambda=0$, the condition on maximal ideals fails: $\mathfrak{m}_{y=0}\mathcal{O}_{X,x=0}=(Y^d)$ which is not even maximal.
For example d), the map is etale precisely because the locus of points where 1) or 3) failed was removed.
Best Answer
Since $a_{ij}(z)$ invertible on the whole of $W$, the determinant of the matrix $(a_{ij})$ is a unit in $A'$. The relations $t_j|_W=\sum_i a_{ij} g_i$ then imply that $t_1|_W,\ldots,t_n|_W$ generate the same ideal as the $g_1,\ldots,g_n$ do.
You need that the elements $t_i|_W$ generate the ideal sheaf for being able to conclude that $f|_U$ is actually the base change of the open immersion $W\hookrightarrow \mathbb{A}^n_S$ along the zero section, which on global sections corresponds to the $R$-algebra morphism $s_0\colon R[t_1,\ldots,t_n]\to R$ sending $t_i$ to $0$.
On the level of rings you can argue as follows: Let $I$ denote the ideal of $A'$ cutting out $U$, i.e. $A=A'/I$. Then you have to show that the diagram involving the zero section $s_0$, the morphism $f|_U$ (on global sections), the projection $A\to A/I=A'$ and the morphism $R[t_1,\ldots,t_n]\to A'$ corresponding to the open immersion $W\hookrightarrow \mathbb{A}^n_S$ is a pushout in the category of rings. Explicitly, given ring morphisms $\varphi\colon A'\to C, \Phi\colon R\to C$ which coincide when being composed with the maps $R[t_1,\ldots,t_n]\to R$ and $R[t_1,\ldots,t_n]\to A'$, respectively, then you need that $I\subset \operatorname{ker}(\varphi)$. But since $s_0$ maps $t_1,\ldots,t_n$ to zero in $R$, you conclude that $\varphi$ maps $t_1|_W,\ldots,t_n|_W$ to zero in $A/I$. Since these generate the ideal $I$, you get the desired inclusion $I\subset \operatorname{ker}(\varphi)$.