Use a for loop

public static class Proverb
{
    public static string[] Recite(string[] subjects)
    {
        if (subjects.Length == 0)
            return Array.Empty<string>();

        var lines = new string[subjects.Length];

        for (int i = 0; i < subjects.Length - 1; i++)
            lines[i] = $"For want of a {subjects[i]} the {subjects[i + 1]} was lost.";

        lines[^1] = $"And all for the want of a {subjects[0]}.";

        return lines;
    }
}

This approach uses a for loop to iterate over the subjects and compose the song's lines.

Song structure

Before we start writing code, let's examine the structure of the song's lines we'll need to return. There are three different situations we have to handle:

1. No subjects: return an empty array
2. One subject: return an array with one element: "And all for the want of a <SUBJECT>."
3. Multiple subjects: return an array of length equal to the number of subjects. All but the last line are of the form: "For want of a <SUBJECT> the <NEXT_SUBJECT> was lost.", where "<SUBJECT>" and "<NEXT_SUBJECT>" are adjacent subjects (subjects at index 0 and 1, then 1 and 2, etc.). The last line is of the form "And all for the want of a <FIRST_SUBJECT>.", where "<FIRST_SUBJECT>" is the first subject.

Subjects: none

Let's check the number of subjects via the array's Length property. If the length is equal to 0, return an empty array:

if (subjects.Length == 0)
    return Array.Empty<string>();

Subjects: one

To implement the case for one subject, we can just return an array with one element that has the correct structure:

return new[] { $"And all for the want of a {subjects[0]}." };

Subjects: multiple

Handling multiple subjects is where the fun begins!

As the array we want to return must have a length equal to the number of subjects, let's start by defining this array:

var lines = new string[subjects.Length];

To reiterate, there will be one line in the returned array for each adjacent subject pair.

Let's look at an example, where the subjects array contains "nail", "shoe" and "horse". There are two adjacent subject pairs in this array:

1. "nail" and "shoe"
2. "shoe" and "horse"

These pairs then should result in the following two lines:

1. "For want of a nail the shoe was lost."
2. "For want of a shoe the horse was lost."

So how do we process there adjacent subject pairs? If we think of the pairs as pairs of indexes (0 and 1, 1 and 2), we can see that the starting index of these pairs (0 and 1) is less than the length of the subjects minus one. That translates to:

for (int i = 0; i < subjects.Length - 1; i++)

Within the for loop, we set each line using its correct value, using the subject at index i and its adjacent subject at index i + 1:

for (int i = 0; i < subjects.Length - 1; i++)
    lines[i] = $"For want of a {subjects[i]} the {subjects[i + 1]} was lost.";

And all for the want of a ...

The last line is of the form: "And all for the want of a <FIRST_SUBJECT>.", which translates to:

lines[lines.Length - 1] = $"And all for the want of a {subjects[0]}."

The last line's index is calculated via lines.Length - 1, but this is quite ugly, so is there a more elegant way? Why yes, there is, by using the index from end operator (^):

lines[^1] = $"And all for the want of a {subjects[0]}."

Simplification

As this code also works for subjects with one element, we can remove the code we've previously written to handle one subject.

Putting it all together

The final step is to return the lines, giving us:

if (subjects.Length == 0)
    return Array.Empty<string>();

var lines = new string[subjects.Length];

for (int i = 0; i < subjects.Length - 1; i++)
    lines[i] = $"For want of a {subjects[i]} the {subjects[i + 1]} was lost.";

lines[^1] = $"And all for the want of a {subjects[0]}.";

return lines;