Isn't that something that would require linear relationships?
The assumption is that the effect of covariates is linear on the log odds scale. You might see logistic regression written as
$$ \operatorname{logit}(p) = X \beta $$
Here, $\operatorname{logit}(p) = \log\left( \frac{p}{1-p} \right)$. Additionally, remember that linearity does not mean straight lines in GLM.
Regarding the errors, the normality assumption isn't required because the errors will be zero or 1?
Not quite. Logistic regression estimates a probability, the error (meaning observation minus prediction) will be between 0 and 1.
Why doesn't Logistic Regression require the error and linear relationship assumptions that Linear Regression require?
Logistic regression is still a linear model, it is just linear in a different space so as to respect the constraint that $0 \leq p \leq 1$. AS for your titular question regarding the error term and its variance, note that a binomial random variable's variance depends on its mean ($\operatorname{Var}(X) = np(1-p)$). Hence, the variance chances as the mean changes, meaning the variance is (technically) heteroskedastic (i.e. non-constant, or at the very least changes based on what $X$ is because $p$ changes based on $X$).
Best Answer
Note that there are times where least squares is applied to modelling where $E(Y)\propto \frac{\exp(X\beta)}{1+\exp(X\beta)}$ ... nonlinear least squares is used for such cases.
However, that case is not the same as logistic regression. One reason we don't apply plain least squares to logistic regression is that the variance of a binomial proportion varies with the proportion -- it's larger when the proportion is $\frac12$ (i.e. when $X\beta$ is 0) than when it's near the extremes.
[So why not weighted least squares? Note that the mean and variance are connected; the variance estimate depends on the estimate of the mean, but the current estimate of the mean depends on the variance ... leading to the need to reweight the least-squares model iteratively -- and that can indeed be done. However, it wouldn't generally be maximum likelihood, which is usually what's desired. A variation on such a scheme, however, can be used to get MLEs.]