Question 1: "So I am wondering if there is an intuitive example to get non-reduced 'schemes' from reduced 'algebraic varieties' (probably by taking fiber product or alike)?"

Nilpotent elements are useful in mathematics (algebra/geometry, complex analysis, differential geometry) when studying derivatives, differential operators and tangent spaces.

Let $k$ be a field and let $A:=k[x],B:=k[x,y]$ be polynomial rings with $I:=(y-x)\subseteq B$ the ideal defined by the element $y-x$. Let $J(l):=B/I^{l+1}$ and let
$T^l:A \rightarrow J(l)$ be defined by $T^l(f(x)):=f(y)$. The left $A$-module $J(l)$
is free of rank $l+1$ on the elements $(dx)^i:=(y-x)^i$ for $i=0,..,l$. You can prove this using induction.

Example 0. Let $k$ have characteristic zero. It follows from the binomial theorem and an induction that

$T^l(f(x))=\sum_{i=0}^l \frac{f^{(i)}(x)}{i!}(dx)^i$.

Example 1. Let $f(x):=x^2$. We get

$T^l(f(x)):=f(y)=y^2=(x+dx)^2=x^2+2xdx+(dx)^2=$

$f(x)+ f'(x)dx+\frac{f^{(2)}(x)}{2!}(dx)^2$.

Hence the map $T^l$ is the Taylor expansion of the polynomial $f(x)$. The map
$T^l$ is a "differential operator" of order $l$ from $A$ to $J(l)$. If $k$ is the field of real numbers and $\mathfrak{m}:=(x-a)$ for $a\in k$ we may pass to the fiber

$J(l)(a):= \kappa(\mathfrak{m})\otimes_A J(l) \cong k\{(dx)^i\}$

and as an element of $J(l)(a)$ it follows

$T^l(f(x)) = \sum_{i=0}^l \frac{f^{(i)}(a)}{i!}(dx)^i \in J(l)(a)$.

Hence in this case we get the value of the Taylor series of $f(x)$ of order $l$ at the
real number $a\in k$.

If $z=a+ib$ with $a,b$ real numbers and $b \neq 0$ it follows

$p(x):=(x-z)(x-\overline{z})=x^2-2ax+a^2+b^2 \in k[x]$

is an irreducible polynomial with $k[x]/(p(x))\cong \mathbb{C}$. Choosing an explicit isomorphism $\phi: k[x]/((p(x)) \cong \mathbb{C}$ it follows

$0=\phi(p(x))=p(\phi(x))$

hence $\phi(x)$ equals $z$ or $\overline{z}$. Hence if $\mathfrak{p}=(p(x))$ and we pass to the fiber $\kappa(\mathfrak{p})\otimes J(l):=J(l)(\mathfrak{p})$ we get an isomorphism

$J(l)(\mathfrak{p})\cong \mathbb{C}\{(dx)^i\}$.

If we "Taylor expand" $f(x)$ and pass to the fiber $J(l)(\mathfrak{p})$

we get an element $T^l(f(\phi(x))) \in \mathbb{C}\{(dx)^i\}$, and the value of the Taylor series $T^l(f(\phi(x)))$

depends on the choice of an isomorphism $\kappa(\mathfrak{p})\cong \mathbb{C}$. and the choice of $\phi(x)\in \mathbb{C}$. What does not depend on this choice is if the higher order derivative $\frac{f^{(i)}(x)}{i!}$ is zero or not.

Note: If $f(x) \in \mathbb{R}[x]$ is any polynomial it follows $T^l(f(x))=0$ in the fiber at $x-a$ iff $f^{(i)}(a)=0$ for $i=0,..,l$. This holds iff $f(x)=f_l(x)(x-a)^{l+1}$. Hence $T^l(f(x))=0$ in the fiber at $x-a$ iff $a$ is a zero of $f(x)$ of order $l+1$. If $p(x):=(x-z)(x-\overline{z})$ it follows $T^l(f(x))=0$ in the fiber at $p(x)$ iff $f(x)=f_l(x)p(x)^{l+1}$. Hence $T^l(f(x))=0$ iff $z$ and $\overline{z}$ are zeros of $f(x)$ of multiplicity $l+1$.

If $\pi: Spec(J(l))\rightarrow Spec(A)$ is the canonial map, it follows $Spec(J(l))$ is a "geometric vector bundle" in the sense of Hartshorne, Ex II.5.18. The scheme $Spec(J(l))$ is non-reduced but $Spec(A)\cong \mathbb{A}^1_k$ is reduced. This gives an intuitive explanation of why non-reduced schemes are important in the study of reduced schemes, answering Question 1. The construction of $J(l)$ can be done in the "language of algebraic varieties" using coherent sheaves.

This construction globalize to any scheme/differentiable manifold/complex manifold/etc.
Hence nilpotent elements and non-reduced ideals are useful in algebraic geometry, differential geometry and complex analysis. In these fields we study tangent spaces and differential operators.

For any smooth projective curve $C \subseteq \mathbb{P}^n_k$ there is a similar construction of a finite rank vector bundle $\mathcal{J}_C^l$ and a universal differential operator

$T^l: \mathcal{O}_C \rightarrow \mathcal{J}_C^l$ which locally "Taylor expands" sections: For any open subset $U \subseteq C$ we get a differential operator

$T^l(U): \mathcal{O}_C(U) \rightarrow \mathcal{J}_C^l(U)$

and the map $T^l(U)$ Taylor expands the section $s\in \mathcal{O}_C(U)$. If $k$ is the field of real or complex numbers, we may view $s$ as a real valued or complex valued function, and in this case $T^l(U)(s)$ is the Taylor expansion of $s$.

Example 2. If $C$ is a complex holomorphic curve and $\mathcal{O}_C$ is the sheaf of holomorphic functions on $C$, there is a similar construction. There is a finite rank holomorphic vector bundle $\mathcal{J}_C^l$ and a differential operator

$T^l: \mathcal{O}_C \rightarrow \mathcal{J}_C^l$

with the same properties.

Example 3. Complex holomorphic curves and nilpotent ideals. The holomorphic vector bundle $\mathcal{J}_C^l$ may be constructed using the ideal of the diagonal $\mathcal{I} \subseteq \mathcal{O}_{C \times C}$. The ideal $\mathcal{I}$ is a coherent sheaf of ideals, and one may prove that the quotient $\mathcal{O}_{C \times C}/\mathcal{I}^{l+1}$ is (as a left $\mathcal{O}_C$-module) locally trivial of rank $l+1$. The corresponding holomorphic vector bundle is isomorphic to $\mathcal{J}_C^l$. The pair $(C\times C, \mathcal{O}_{C\times C}/\mathcal{I}^{l+1})$ may be viewed as a "locally ringed space" with nilpotent elements in the structure sheaf. The sheaf $\mathcal{J}_C^l$ is supported on the diagonal $\Delta(C) \subseteq C \times C$, and it follows $\mathcal{J}_C^l$ is a sheaf of left and right $\mathcal{O}_C$-modules. The sheaf $\mathcal{J}_C^l$ is locally trivial as left and right $\mathcal{O}_C$-module, but these two structures are not isomorphic in general. If you consider the ideal sheaf $\overline{\mathcal{I}} \subseteq \mathcal{J}_C^l$ and choose any non-zero section $s \in \overline{\mathcal{I}}(U)$ for $U \subseteq C$ an open set it follows $s^{l+1}=0$. Hence the sheaf of rings $\mathcal{J}_C^{l}$ has nilpotent elements. When we restrict $\mathcal{J}_C^{l}$ to the diagonal $\Delta(C)$
we get a sheaf of left and right $\mathcal{O}_C$-modules, but for local sections $s \in \mathcal{O}_C(U)$ and $\omega \in \mathcal{J}_C^l(U)$ it follows $s\omega \neq \omega s$.

You may have seen in Hartshorne exercise II.5.18 that for a scheme $(X,\mathcal{O}_X)$ there is an "equivalence of categories" between the category of locally free finite rank $\mathcal{O}_X$-modules and the category of finite rank geometric vector bundles, and there is a similar result valid for complex holomorphic manifolds. Given a complex holomorphic manifold $(Y, \mathcal{O})$ where $\mathcal{O}$ is the sheaf of complex holomorphic functions
on $Y$, it follows there is an "equivalence of categories" between the category of locally trivial finite rank $\mathcal{O}$-modules and the category of finite rank holomorphic vector bundles on $Y$.

In a calculus course in one variable you define the derivative $f'(x)$ of a real valued smooth function $f(x)$ using limits. In algebraic geometry we work over arbitrary fields (or rings) and we cannot "take limits". Using non-reduced ideals we can still formally "take derivatives" of sections of sheaves as explained above. In characteristic zero we get as you can observe above the correct Taylor expansion of any polynomial (or rational) function. In characteristic $p>0$ the notion is useful as well, ref. Question 1 in your question.

See also the discussion at:

https://math.stackexchange.com/questions/3928532/about-the-reducedness-in-algebraic-geometry/3942701#3942701

howthey generalize varieties outside of the very formal passage from polynomials rings to general rings and a vague statement about generic points. It's not clear to me how much detail or motivation an encyclopedia should carry, but along with "generalized points," nilpotents are one of the hallmarks of the passage from varieties to schemes, and certainly deserve mention. $\endgroup$ – Ramsey Feb 12 '11 at 23:24