Let $x = (x_1, x_2, \ldots, x_n)$ be the series. Set
$$y_t = x_t - \bar{x}.$$
These are the residuals with respect to the estimated mean $\bar{x} = \frac{1}{n}\sum_{t=1}^n x_t$ of the series.
For $k=0, 1, 2, \ldots, n-1$ the acf function is computing
$$\text{acf}(x)_k = \frac{\sum_{t=1}^{n-k} y_t y_{t+k}}{\sum_{t=1}^n y_t^2}.$$
Notice that as the lag $k$ grows, there are fewer and fewer terms in the numerator as well as a shift of the indexes in the product. The reduction in number of terms in the numerator essentially forces a decrease in the value as $k$ increases. Most time series analyses consider only lags $k$ much smaller than $n$ for which this effect is negligible.
In your example where $x = (0, 1, 2, 3, 4, 5)$, $y = (-5/2, -3/5, -1/2, 1/2, 3/2, 5/2)$ initially has negative values and then moves into positive territory. For lags $k \ge 3$, the products $y_ty_{t+k}$ are pairing the early negative values with the later positive values, producing negative numbers.
Edit: Intuitive Explanation
Intuitively, $\text{acf}(x)_k$ is supposed to be telling us the correlation between a series and its lag-$k$ version. The motivation for the question is that a series like $(0, 1, \ldots, n-1)$ is perfectly correlated with all its lags for $k=0$ right through $k=n-2$. How, then, can the ACF plot produce near zero and even negative values?
There are two factors in play here. They can be seen by comparing the ACF formula to that of the usual correlation coefficient. For two series $(u_t)$ and $(w_t)$ of the same length $n-k$, let $\upsilon_t = u_t - \bar{u}$ and $\omega_t = w_t - \bar{w}$ be their residuals. (In the ensuing discussion, $(u_t)$ will be the prefix $(x_1, x_2, \ldots, x_{n-k}$ and $(w_t)$ will be the suffix $(x_{k+1}, x_{k+2}, \ldots, x_n)$.) By definition, their correlation coefficient is the average standardized residual,
$$\rho(u, w) = \frac{\sum_{t=1}^{n-k} \upsilon_t \omega_t}{\sqrt{\sum_{t=1}^{n-k} \upsilon_t^2 \sum_{t=1}^{n-k} \omega_t^2}}.$$
(The constants $\frac{1}{n-k}$ that usually appear in formulas for averages cancel in this ratio, so I have omitted them.)
When we are dealing with a single series $(x_t)$ of length $n$ and its (short) lags $k$, both $\upsilon_t$ and $\omega_t$ are essentially the same, apart from the shift of $k$ in their indexes: the first consists of the $(y_t)$ for $t$ from $1$ through $n-k$ (the high-$t$ end has been trimmed off) while the second consists of the same $(y_t)$ for $t$ from $k$ through $n$ (the low-$t$ end has been removed). If we ignore these slight differences, the denominator of $\rho(u, w)$ simplifies to
$$\sqrt{\sum_{t=1}^{n-k} \upsilon_t^2 \sum_{t=1}^{n-k} \omega_t^2} = \sqrt{\sum_{t=1}^{n-k} y_t^2 \sum_{t=1}^{n-k} y_{t+k}^2} \approx \sqrt{\sum_{t=1}^{n} y_t^2 \sum_{t=1}^n y_{t}^2} = \sqrt{\left(\sum_{t=1}^{n} y_t^2\right)^2 } = \sum_{t=1}^{n} y_t^2.$$
In making this approximation I have inserted the first $k$ terms $y_1^2 + \cdots + y_k^2$ into the sum for the suffix ($\omega_t$) and the last $k$ terms $y_{n-k+1}^2 + \cdots + y_{n}^2$ into the sum for the prefix ($\upsilon_t$). Because these are both sums of squares, they cannot decrease the denominator, and usually increase it a little bit. Accordingly, we see that using $\sum_{t=1}^n y_t^2$ in the denominator decreases the apparent correlation $\rho(u, w)$. The greater the lag $k$, the more the denominator will tend to increase, so this factor tends to reduce the high-lag values of the ACF no matter what.
The second factor has to do with the difference between the mean of the entire series $\bar{x}$ and the means of the prefix $\bar{\upsilon} = \frac{1}{n-k}\sum_{t=1}^{n-k} y_t$ and suffix $\bar{\omega} = \frac{1}{n-k}\sum_{t=k+1}^n y_t$. The ACF formula uses the former whereas the correlation coefficient formula uses the latter. We can work out the change in the numerator by comparing the ACF and correlation coefficient formulas, working algebraically to make the ACF numerator look like the $\rho$ numerator:
$$\eqalign{
\sum_{t=1}^{n-k} y_t y_{t+k} &= &\sum_{t=1}^{n-k} (x_t-\bar{x})(x_{t+k}-\bar{x}) \\
&= &\sum_{t=1}^{n-k} (x_t-\bar u + \bar u - \bar{x})(x_{t+k}-\bar w + \bar w - \bar{x}) \\
&= &\sum_{t=1}^{n-k} \left((x_t-\bar u)(x_{t+k}-\bar w) + (\bar u - \bar{x})(\bar w - \bar{x})\right) \\
&= &\left(\sum_{t=1}^{n-k} \upsilon_t \omega_t\right) + (n-k)(\bar u - \bar{x})(\bar w - \bar{x}).
}$$
(The cross terms disappeared after the second line for the usual reason: they sum to zero.)
Comparing to the formula for $\rho$, we see that the discrepancy in numerators depends on the lag (in terms of $n-k$) and the products of the changes in the means, $\bar u - \bar{x}$ and $\bar w - \bar{x}$. For a stationary series and large $k$ those changes ought to be small; for small $k$ we hope they will be small but perhaps not. In the example, for instance, at lag $k=1$ the mean after dropping off the last term decreases by $1/2$ and the mean after dropping off the first term similarly increases by $1/2$. The product
$$(n-k)(\bar u - \bar{x})(\bar w - \bar{x}) = (6-1)(-1/2)(1/2) = -5/4$$
decreases the numerator in the ACF compared to the numerator in $\rho$.
The net effect of these two factors in the example is that both conspire to decrease the apparent correlation: the denominator goes up, because it includes a few more positive terms overall, and the numerator goes down, because one end of the series tends to be less than the average and the other end tends to be greater than the average. (That's more or less what a "long term trend" means, suggesting there is some evidence of non stationarity in this series.)
To illustrate the formula for the ACF, here is direct (but less efficient) R code to compute acf:
acf.0 <- function(x) {
n <- length(x)
y <- x - mean(x)
sapply(1:n - 1, function(k) sum( y[1:(n-k)] * y[1:(n-k) + k] )) / sum(y * y)
}
As a test, compare the two results:
> sum((acf.0(0:5) - acf(0:5, plot=FALSE)$acf)^2)
> 6.162976e-33
The answers agree to within double precision floating point roundoff error.
Best Answer
Sample ACF is not decreasing linearly. You will get the idea of the pattern by calculating corr($S_n, S_m$)
The answer is followed from the definition of PACF. The above random walk you can express as follows. $$S_n = S_{n-1} +X_n \qquad n=1,2,\ldots,N, \text{ where } S_0=0 $$
You can take the second PACF expression as correlation between
$$S_{t+2} − \mathcal P (S_{t+2} \mid S_{t+1}) \text{ and } S_{t} − \mathcal P (S_t \mid S_{t+1} ),$$ where $ \mathcal P(W\mid Z)$ is the best linear projection of $ W$ on $ Z$.
$S_{t+2} − \mathcal P (S_{t+2} \mid S_{t+1}) =X_{t+2}$ and second term will be linear function of $X_{i}, i=1,2,\ldots,t+1 $. Since $X_{n}$'s are iid hence, correlation (second order PACF) is zero. Similar way you could prove higher order PACF are zeros.