On Wikipedia it says:
Clifford algebras are closely related to exterior algebras. Indeed, if $Q$ = 0 then the Clifford algebra $Cl(V, Q)$ is just the exterior algebra ⋀$V$.
With $Q$ being a quadratic form $Q : V → K$. Is it correct to think of this as a scalar product (for example or what else?), such that this sentence above then just means that all $v \in V$ are orthogonal?
Only if the characteristic is not 2. In this case we get the bilinear form $$ B(u,v) = Q(u+v) - Q(u) - Q(v) $$ and $Q(u) = 0$ for all $u$ is equivalent to $B(u,v) = 0$ for all $u, v$.
However, in characteristic 2 the diagonal of $B$ is always trivial $$ B(u,u) = Q(2u) - Q(u) - Q(u) = Q(0) - 2Q(u) = 0 - 0 = 0. $$ What this means is that the theory of bilinear forms and the theory of quadratic forms are not equivalent in characteristic 2. So while it's reasonable to say that "$Q = 0$ implies all vectors are orthogonal to each other", the converse doesn't make much sense.
For a concrete example, consider $B'$ over $\mathbb F_2^2$ defined by $$ B'(u,v) = u^T\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}v. $$ Then we get a quadratic form $Q(u) = B'(u,u)$ which is nonzero since $(1,0)^T \mapsto 1$, but the bilinear form $B$ associated with $Q$ is trivial: $$ B(u,v) = Q(u+v) - Q(u) - Q(v) = B'(u+v,u+v) - B'(u,u) - B'(v,v) = B'(u,v) + B'(v,u) = 2B'(u,v) = 0 $$