Let $\alpha$ be a given positive real number. We aim to find all functions $f: \mathbb{N}^{+} \rightarrow \mathbb{R}$ such that $f(k + m) = f(k) + f(m)$ holds for any positive integers $k$ and $m$ satisfying $\alpha m \leq k \leq (\alpha + 1)m$.

To solve this, we first note that the given functional equation $f(k + m) = f(k) + f(m)$ resembles Cauchy's functional equation. However, the condition $\alpha m \leq k \leq (\alpha + 1)m$ restricts the values of $k$ and $m$.

We will show that the only solution to this functional equation under the given condition is a linear function of the form $f(n) = cn$ for some constant $c$.

1. Step 1: Prove separability for large integers

An integer $n$ is called separable if there exist integers $a$ and $b$ such that $a + b = n$ and $f(a + b) = f(a) + f(b)$. We need to show that all sufficiently large integers are separable.

Given the condition $\alpha m \leq k \leq (\alpha + 1)m$, we can rewrite it in terms of $n = k + m$ as:
$$\alpha + 1 \leq \frac{n}{m} \leq \alpha + 2.$$
This implies:
$$\frac{1}{\alpha + 2} \leq \frac{m}{n} \leq \frac{1}{\alpha + 1}.$$
By starting with $\frac{1}{n}$ and adding $\frac{1}{n}$ at each step, we ensure that $m$ remains within the interval $\left( \frac{1}{\alpha + 2}, \frac{1}{\alpha + 1} \right)$. Hence, such an integer $n$ is separable.

2. **Step 2: Represent $f(n)$ as a linear combination**

Since all sufficiently large integers are separable, $f(n)$ can be expressed as a linear combination of $f(1), f(2), \ldots, f(k)$ for some fixed $k$. Let us consider the smallest subset such that the linear representation is unique:
$$f(n) = g_1(n)f(a_1) + \cdots + g_m(n)f(a_m),$$
where $\{a_1, a_2, \ldots, a_m\} \subset \{1, 2, \ldots, k\}$.

By the uniqueness of the linear representation, the problem condition carries over to the $g_i$ functions. Since $g_i$ are functions from integers to rationals, and every rational can be represented as a linear combination of other rationals, it follows that $g_i$ must be linear functions of the form $g_i(n) = n c_i$ for some constants $c_i$.

3. **Step 3: Conclude the form of $f(n)$**

Therefore, we can represent $f$ as:
$$f(n) = n (c_1 f(a_1) + \cdots + c_m f(a_m)) = nD,$$
where $D$ is a constant. This representation holds for all $n > k$.

4. **Step 4: Verify for all $n$**

For $\alpha \leq 1$, we have $f(2n) = 2f(n)$. By induction, choosing a sufficiently large $k$, we get $f(n) = nD$.

For $\alpha > 1$, choose an integer $a$ such that $\alpha \leq a \leq \alpha + 1$. We have $f((a+1)n) = f(n) + f(an)$. By choosing $n$ in the interval $(k/2, k]$, we extend $f(n) = nD$ to all $n$.

Thus, the only function $f$ satisfying the given conditions is:
$$f(n) = cn,$$
where $c$ is a constant.

The answer is: \boxed{f(n) = cn}.