The fourth property is called "positive definiteness". Typically when moving over to other fields one replaces this with "non-degeneracy" (which is weaker than positive definiteness).
4'. $\langle v,w \rangle=0$ for all $w$ implies that $v=0$.
Note: If you move to working over the complex numbers then usually one replaces 3. with conjugate symmetry. And so this along with 2. imply that pulling scalars out of the second slot results in conjugation.
An inner product is an additional structure on a vector space. It is true that all vector spaces have inner products, but there can be two different inner products on the same vector space. For instance, on $\mathbb{R}^2$ one has the "usual" inner product $\langle a,b \rangle \cdot \langle c,d \rangle= ac+bd$. But if you change coordinates by replacing the "standard basis" $\langle 1,0 \rangle$, $\langle 0,1 \rangle$ by some other basis then you may get other inner products.
The additional structure that an inner products gives to a vector space is geometric in nature. First, the inner product gives a way of measuring lengths of vectors, using the formula
$$Length(v) = \sqrt{v \cdot v}
$$
Second, the inner product gives a way of measuring angles between two vectors, using the law of cosines formula
$$Angle(v,w) = \cos^{-1}[(v \cdot w) / (Length(v) \, Length(w))]
$$
If you change inner products on the same vector space, then you may get two different angle measurements, two different length measurements, two different notions of "circles", etc. Just as an example, if you define an inner product on $\mathbb{R}^2$ using the basis $\langle 2,0 \rangle$, $\langle 0,1 \rangle$, then the "circles" in this geometry are ellipses whose major axis along the $x$-axis has twice the ordinary Euclidean length as their minor axis along the $y$-axis.
Best Answer
Sure. Let $V=\mathbb R^3$ be a real vector space with standard basis $e_1,e_2,e_3$. Define $f:V \to \mathbb R^3$ by $(e_1,e_2,e_3) \to (e_1,e_2,2e_3)$ and take the standard inner product (dot product) in $\mathbb R^3$ $\langle \cdot, \cdot \rangle_3$ for $\mathbb R^3$. Then, we can define $\langle x,y \rangle_*:= \langle f(x),f(y) \rangle_3$.
Note that $(1,0,1)$ and $(4,0,-1)$ are orthogonal in the inner product with subscript $*,$ i.e., $\langle \cdot,\cdot\rangle_*,$ but not in the standard inner product. Likewise, $(1,1,-2)$ and $(1,1,1)$ are orthogonal in the standard inner product, but not in the inner product with subscript $*.$