For two random variables $X$ and $Y$ we know that the following holds regarding the variance
$$ \text{Var}(X \pm Y) = \text{Var}(X) + \text{Var}(Y) $$
We also know from the definition of Brownian motion $B$ that
$$ B(t+s)-B(t) \sim N(0, s) $$
These two statements clearly contradict each other since the former adds the variances while the latter subtracts them.
Why is that? A Brownian motion is also a random varible, so why doesn't it satisfy the first relation above?
Your formula assumes that $X$ and $Y$ are independent. The general formula for non independent r.v. is $$var(X-Y)=var(X)+var(Y)-2cov(X,Y).$$ Since $cov(B(t),B(t+s))=\min(t+s,t)=t$ we have that $$var(B(t+s)-B(t))=t+s+t-2t =s.$$