Since this is a homework problem, I will beat around the bush a bit and will not be giving a final determination about how to interpret your $R^2$ score. Hopefully, however, you can use what I write here to inform your homework submission.
Also the average of the residuals is not 0. It is -2.8422E-15, so close to 0, but not really actually 0.
There is no assumption about the residuals having a mean of zero. Residuals having a mean of zero in least squares linear regression is a mathematical fact. When you use a linear model (with an intercept) and compute the coefficients by finding the coefficients that minimize the sum of the squared residuals ("least squares"), it is a mathematical fact that the residuals will have a mean of zero.
As you point out, your residuals do not have a mean of exactly zero but of $-0.0000000000000028422$, very close to, but not exactly equal to, zero. Thus, your estimates are not the exact least squares estimates (I assume your Excel implementation includes an intercept, which I find to be a safe assumption).
However, the mean is so close to zero that this is pretty much irrelevant. The true least squares values probably have many more decimal digits then the computer is tracking. In most cases, I would say those do not matter, so your residual mean of -2.8422E-15 is (probably) safely assumed to be "basically zero".
Really, the residual mean being zero is more of a test of if the given solution really is a least squares solution than it is about any statistical assumptions, since the residuals having a mean of zero is a property of the least squares solution, whether a least squares approach is wise or not.
But my notes also say that linear regression is not the correct model if the residuals are not distributed normally
This is a dubious claim. For instance, while this might not be part of your course, there is this Gauss-Markov theorem that proves a nice property of the least squares estimator, and there is no reference to normality in the setup of the Gauss-Markov theorem. When you have normal errors (not quite the same as residuals, but you can use the residuals to gauge or estimate the errors) with a constant variance, what you get is that the least squares estimator coincides with maximum likelihood estimation, which implies some desirable properties and some typical calculations for coefficient confidence intervals and hypothesis tests that are not quite as accurate when the error normality is violated. However, that has nothing to do with the coefficient of determination.
Finally, it is even a mistake to conflate linear regression with anything involving estimation technique or error distribution. A linear model can be totally valid for a different estimation technique than square loss when you assume a different distribution for the errors, such as minimizing absolute loss when the errors are assumed to be Laplace-distributed (which turns out to be equivalent to maximum likelihood estimation). However, you still have the same $\mathbb E\left[y\right] = X\beta$ linear regression formula, just a different way to estimate $\beta$. Likewise, linear regression would be inappropriate for a situation like the following, where the curve is specified to follow $\mathbb E\left[y\right] = \exp\left(\beta_0 + \beta_1x\right)$.
However, the residuals from fitting the correct (nonlinear) regression are quite normal, as the normal quantile-quantile plot below shows.
R Code:
set.seed(2024)
N <- 10000
x <- runif(N, -2, 2)
z <- 2 + x
u <- exp(z)
y <- u + rnorm(N, 0, 1)
plot(x, y)
L <- glm(y ~ x, family = gaussian(link = "log"), start = c(2, 1))
r <- resid(L)
qqnorm(r)
qqline(r)
