I’ve learned several key and practical points that have perfectly resolved my confusion. First, I’ve grasped the core trick of disentangling a complex piecewise function into individual piecewise components using the pattern-matching method

pwApart = HoldPattern@Piecewise[lis_, 0] :> (Piecewise[{#}, 0] & /@ lis)

This approach breaks down the original complicated piecewise general term into simpler, separate parts, which makes the subsequent summation operation much more manageable and avoids the messy results caused by directly summing the combined piecewise function.
Second, I’ve understood the importance of the combination of FunctionExpand, Total and Simplify functions in post-processing the summation result. These functions work together to transform the initial DifferenceRoot form into a concise algebraic expression, which is exactly the explicit form I was eager to obtain.
Third, your guidance on converting the (-1)^n term to a piecewise function is incredibly useful. The method of using Assuming to set the domain constraint of positive integers, then replacing (-1)^n with a piecewise expression and applying PiecewiseExpand helps to present the summation result in a more intuitive piecewise form that aligns with the parity segmentation feature of the original sequence. I also noticed your reminder about the default value mechanism of the Piecewise function, which helped me understand why the result format has a slight difference from my initial expectation. Your solution is not only a fix for this specific problem but also provides me with systematic thinking for handling piecewise sequence summation in the future. Thank you again for taking the time to share your expertise and clear up my confusion!

What matters is the correctness of the method applied to this problem. In contrast to the result presented in the figure, we only mandate such a display form.

What is the proper approach when a piecewise form is employed? Is it feasible to apply the Piecewise function to formulate the general term formula of the sequence, thereby guaranteeing the validity of the subsequent summation operation?

Another way:

a[n_] = (5 n + 1) (1 - Mod[n, 2]) + 2^(n/2) Mod[n, 2]
s[n_] = Sum[a[j], {j, 1, n}]
Piecewise[{{Simplify[s[n], Mod[n, 2] == 0], Mod[n, 2] == 0},
  {Simplify[s[n], Mod[n, 2] == 1], Mod[n, 2] == 1}}]

Simpler solution without Piecewise :

a1[n_] = 
  Piecewise[{{5 n + 1, Mod[n, 2] == 0}, {2^(n/2), Mod[n, 2] == 1}}];
c1[n_] = 
 Sum[(a1[n][[1, 1, 1]]) /. n -> 2 k, {k, 1, Floor[n/2]}] + 
   Sum[(a1[n][[1, 2, 1]]) /. n -> 2 k - 1, {k, 1, Ceiling[n/2]}] // 
  Expand

(*-Sqrt[2] + 2^(1/2 + Ceiling[n/2]) + 6 Floor[n/2] + 5 Floor[n/2]^2*)

a1[n_] = 
  Piecewise[{{5 n + 1, Mod[n, 2] == 0}, {2^(n/2), Mod[n, 2] == 1}}];
c1[n_] = 
 Sum[(a1[n][[1, 1, 1]]) /. n -> 2 k, {k, 1, Floor[n/2]}] + 
  Sum[(a1[n][[1, 2, 1]]) /. n -> 2 k - 1, {k, 1, Ceiling[n/2]}]; 
Piecewise[{{Simplify[c1[n], Mod[n, 2] == 0], 
    Mod[n, 2] == 0}, {Simplify[c1[n], Mod[n, 2] == 1], 
    Mod[n, 2] == 1}}] // ExpandAll