Here's an elementary proof of a related inequality (with non-sharp constants), which may explain what Iosif said in his answer. For ease of typing, instead of $\lambda^{-k}$ I will just type $y \ll 1$.

Let $f$ be a $C^1$ function vanishing at $1$, then you have

$$ \int_y^1 (x f(x)^2)' = - y f(y)^2 $$

Expanding the left you get

$$ \int_y^1 f(x)^2 + 2 \int_y^1 x f(x) f'(x) = - y f(y)^2 $$

rearrange and use Cauchy-Schwarz, you get

$$ \int_y^1 f(x)^2 \leq \big(\int_y^1 x^{2-\alpha} f(x)^2\big)^{1/2} \big(\int_y^1 x^\alpha f'(x)^2 \big)^{1/2} + y f(y)^2 $$

Young's inequality makes it, for every $\epsilon > 0$:

$$ \int_y^1 f(x)^2 \leq \epsilon \int_y^1 x^{2-\alpha} f(x)^2 + \frac1{\epsilon}\int_y^1 x^\alpha f'(x)^2 + y f(y)^2 $$

Since $\alpha < 2$, we have $x^{2-\alpha} \leq 1$, so we can absorb

$$ (1-\epsilon) \int_y^1 f(x)^2 \leq \frac{1}{\epsilon}\int_y^1 x^\alpha f'(x)^2 + y f(y)^2 $$

To handle the boundary term, you can write

$$ f(y)^2 = \big( \int_y^1 f'(x) \big)^2 \leq (1-y) \int_y^1 f'(x)^2 \leq (1-y) y^{-\alpha} \int_y^1 x^\alpha f'(x)^2 $$

So you get, all things considered

$$ \int_y^1 f^2 \leq \frac{1}{1-\epsilon} \left( \frac{1}{\epsilon} + y^{1-\alpha}(1-y) \right) \int_y^1 x^\alpha (f')^2$$

Choosing $\epsilon = 1/2$ you get

$$ \int_{\lambda^{-k}}^1 f^2 \leq (4 + 2 \lambda^{-(1-\alpha)k}) \int_{\lambda^{-k}}^1 x^\alpha |f'|^2 \leq 6 \int_{\lambda^{-k}}^1 x^\alpha |f'|^2 $$

So, to make this look like your original expression, if $C$ is allowed to be a function of $\lambda$, you can choose $C = 6 \lambda^p$ and the desired inequality will hold.

But you cannot recast this inequality in the form of $C\lambda^{-p}$ with a constant $C$ independent of $\lambda$, due to the number $4$ that shows up.

Here's an elementary proof of a related inequality (with non-sharp constants), which may explain what Iosif said in his answer. For ease of typing, instead of $\lambda^{-k}$ I will just type $y \ll 1$.

Let $f$ be a $C^1$ function vanishing at $1$, then you have

$$ \int_y^1 (x f(x)^2)' = - y f(y)^2 $$

Expanding the left you get

$$ \int_y^1 f(x)^2 + 2 \int_y^1 x f(x) f'(x) = - y f(y)^2 $$

rearrange and use Cauchy-Schwarz, you get

$$ \int_y^1 f(x)^2 \leq 2\big(\int_y^1 x^{2-\alpha} f(x)^2\big)^{1/2} \big(\int_y^1 x^\alpha f'(x)^2 \big)^{1/2} + y f(y)^2 $$

Young's inequality makes it, for every $\epsilon > 0$:

$$ \int_y^1 f(x)^2 \leq \epsilon \int_y^1 x^{2-\alpha} f(x)^2 + \frac1{\epsilon}\int_y^1 x^\alpha f'(x)^2 + y f(y)^2 $$

Since $\alpha < 2$, we have $x^{2-\alpha} \leq 1$, so we can absorb

$$ (1-\epsilon) \int_y^1 f(x)^2 \leq \frac{1}{\epsilon}\int_y^1 x^\alpha f'(x)^2 + y f(y)^2 $$

To handle the boundary term, you can write

$$ f(y)^2 = \big( \int_y^1 f'(x) \big)^2 \leq (1-y) \int_y^1 f'(x)^2 \leq (1-y) y^{-\alpha} \int_y^1 x^\alpha f'(x)^2 $$

So you get, all things considered

$$ \int_y^1 f^2 \leq \frac{1}{1-\epsilon} \left( \frac{1}{\epsilon} + y^{1-\alpha}(1-y) \right) \int_y^1 x^\alpha (f')^2$$

Choosing $\epsilon = 1/2$ you get

$$ \int_{\lambda^{-k}}^1 f^2 \leq (4 + 2 \lambda^{-(1-\alpha)k}) \int_{\lambda^{-k}}^1 x^\alpha |f'|^2 \leq 6 \int_{\lambda^{-k}}^1 x^\alpha |f'|^2 $$

So, to make this look like your original expression, if $C$ is allowed to be a function of $\lambda$, you can choose $C = 6 \lambda^p$ and the desired inequality will hold.

But you cannot recast this inequality in the form of $C\lambda^{-p}$ with a constant $C$ independent of $\lambda$, due to the number $4$ that shows up.