This is an abuse of notation, but you can always write $Y \,\vert\, X = \mathbb{E}\left[Y\,\vert\, X\right] + \epsilon \,\vert\, X$ where $\mathbb{E}\left[\epsilon \,\vert\, X\right]$ is necessarily zero. (If it were not , you could take expectations on both sides and get $\mathbb{E}\left[Y\,\vert\, X\right] \neq \mathbb{E}\left[Y\,\vert\, X\right]$.)
In words: conditional on X, Y equals its conditional mean plus some mean-zero residual (whose distribution might nonetheless depend on X, e.g. the variance of epsilon might change with X).
I interpret "the dependent variable is linearly related to the independent variables" in your quote as meaning $\mathbb{E}\left[Y\,\vert\, X\right] = \beta^{T}X$, i.e. the mean of $Y$ given $X$ is linear in $X$.
If that assumption is true, the model residuals $\hat{\epsilon} \equiv Y - \hat{\beta}^{T}X$, where $\hat{\beta}$ is your estimate of $\beta$, should have (approximately) zero mean for all values of $X$, and hence for all values of $\hat{\beta}^{T}X$. In other words, you're looking to see whether $\mathbb{E}\left[Y - \beta^{T}X \,\vert\, X\right] = 0$ for all $X$, using the estimated coefficients since you don't know the true $\beta$.
Perhaps the point is made most clearly if I write: $\mathbb{E}\left[Y - \mathbb{E}\left[Y\,\vert\, X\right] \,\vert\, X\right] = 0$ always holds. Linearity of the conditional mean implies $\mathbb{E}\left[Y - \beta^{T}X \,\vert\, X\right] = 0$ for all values of $X$, which implies there is "no systematic relationship between the residuals and the fitted values" -- if you interpret "systematic relationship" as meaning a change in conditional mean.
An example where the linearity assumption is violated: suppose $Y \,\vert\, X = \sin X + \epsilon \,\vert\, X$ where $\epsilon \,\vert\, X$ has mean zero:
n <- 10^3
df <- data.frame(x=runif(n, 1, 10))
df$mean.y.given.x <- sin(df$x)
df$y <- df$mean.y.given.x + rnorm(n)
model <- lm(y ~ x, data=df)
plot(df$x, residuals(model)) # Looks like the mean of the residuals varies with X
plot(predict(model, newdata=df), residuals(model)) # Exact same issue
