difference in spectrum
If I understand correctly, the OP is comparing cascading 5 time domain signals one after the other as in $[f_1, f_2, f_3, f_4, f_5]$ and taking the DFT to summing longer samples of the signals and then taking the DFT. The following is what we would expect to see in this case.

The DFT of the sum of functions is equal to the sum of their DFT's:

$$\sum_{n=0}^{N-1}f_1(n)W_N^{nk}+\sum_{n=0}^{N-1}f_2(n)W_N^{nk} = \sum_{n=0}^{N-1}(f_1(n)+f_2(n))W_N^{nk}$$ 

Consider first all sequences of the same length and cascading versus summing the sequences:

Cascading $R$ sequences of equal length $M$ is identical to zero-padding each of the sequences out to $N = RM$, each with  starting offset of $rM$ where $r$ is from $0$ to $R$.

Example with two sequences with $M=4$:

$f_1 = [1, 2, 3, 4]$

$f_2 = [6, 8, 9, 1]$

$\text{DFT}\{[1, 2, 3, 4, 6, 8, 9, 1]\}$ 

$ =  \text{DFT}\{[1, 2, 3, 4, 0, 0, 0, 0]\} + \text{DFT}\{0,0,0,0, 6, 8, 9, 1]\}$

Zero-padding a sequence prior to taking the DFT does not change the original DFT values, but inserts new interpolated values in between the samples in frequency. Such that

$$F_p(nk) = F(k) \tag{1}\label{1}$$

Where $F$ is the DFT of a sequence $f$ and $F_p$ is the DFT of $f$ with zero-padding. This is proven below along with the fact that equation $\ref{1}$ holds true for all offsets $rM$.   

So in the example with $f_1$ aand $f_2$ above, $$F[k] = \text{DFT}\{[1, 2, 3, 4, 6, 8, 9, 1]\} $$

$$F[nk] = \text{DFT}\{[1, 2, 3, 4]\} + \text{DFT}\{[6, 8, 9, 1]\}$$

Similarly, the DFT of the cascade of the five sequences would have the same values as the DFT of the sum of the five sequences with four interpolated values in between each of the DFT samples!

Since in this case the OP didn't simply sum the sequences, but rather summed longer sequences out to the same length as the cascaded sequence, then here the DFT of the longer sequences would have the higher frequency resolution and scaled 5 times larger given by the longer data set, so the values in between that were previously interpolated would now be dictated by the longer time samples as new DFT values. However every 5th sample in both sets should match exactly other than the scale of 5 for the same reason outlined above, with more samples of actual frequency data in between! 



Proof of relationship used above in equation $\ref{1}$

Given sequence $f(n)$ where $f(n) = 0 \text{ for } n \notin [0, M-1]$

The DFT of $f(n)$ is:
$$F(k)=\sum_{m=0}^{M-1} f(m)W_M^{mk}, \space\space\space\space k = 0,...M-1\tag{2} \label{2}$$

Where $W_M^{n}$ are the $n$ roots of unity given by $e^{-j2\pi n/N}$ 

The zero padded DFT of $f(n)$  out to length $N=RM$ with $R\ge 0$ with any offset $rM$ with $r \in [0, R]$ is:

$$F_P(k) = \sum_{n=0}^{N-1} f(n-rM)W_{N}^{nk}, \space\space\space\space k = 0,...N-1\tag{3} \label{3}$$

Using the DFT shift theorem, equation $\ref{3}$ becomes:

$$F_P(k) = \sum_{n=0}^{RM-1} f(n-rM)W_{RM}^{nk}=W_{RM}^{rMk}\sum_{n=0}^{RM-1} f(n)W_{RM}^{nk} $$
$$= W_{R}^{rk}\sum_{n=0}^{RM-1} f(n)W_{RM}^{nk}$$

Since $f(n) = 0$ for $n&gt;M-1$

$$F_P(k) = W_{R}^{rk}\sum_{n=0}^{M-1} f(n)W_{RM}^{nk}\tag{4} \label{4}$$

$$F_P(Rk) = W_{R}^{rRk}\sum_{n=0}^{M-1} f(n)W_{RM}^{nRk} = \sum_{n=0}^{M-1} f(n)W_{M}^{nk} = F(k)$$

Thus we see how given any sequence $f(n)$ of length $M$ with DFT $F(k)$, zero padded out to length $N=RM$ for any positive integer $R$ with DFT $F_P(k)$, the following equality will hold for any rotational offset (starting index) of $rM$ for any integer $r$ from $0$ to $R-1$:

$$F(k) = F_p(Rk) \tag{5} \label {5}$$