Since the computation is essentially the same, let's do it in the $2$-dimensional case. So let us assume $X=\mathbb{A}^2$ with coordinates $x, y$. We have

$\tilde{X}=\{\lambda x+ \mu y=0\} \subset \mathbb{A}^2 \times \mathbb{P}^1$.

In the affine chart $\mu=1$ the equation is therefore $y=\lambda x$, and the exceptional divisor $E$ is given by $x=0$.

Now let $\omega$ be a meromorphic $2$-form on $X$, that we can write as

$\omega=\frac{f(x,y)}{g(x,y)}dx \wedge dy$.

Therefore

$\sigma^* \omega=\frac{f(x, \lambda x)}{g(x, \lambda x)} dx \wedge d(\lambda x)= x \frac{f(x, \lambda x)}{g(x, \lambda x)} dx \wedge d \lambda$.

Therefore in a neighborhood of $E$ we have

$\textrm{div} \ (\sigma^* \omega)= \sigma^* (\textrm{div} \ \omega) + E$.

Since clearly $\textrm{div} \ (\sigma^* \omega)= \sigma^* (\textrm{div} \ \omega)$ outside $E$, it follows in particular that the divisor of every holomorphic form on $\tilde{X}$ contains the component $x=0$, i.e. the exceptional divisor. In other words, the exceptional divisor is contained in the fixed part of the canonical system of $\tilde{X}$, namely

$|K_{\tilde{X}}|= E + |K_X|$.

This implies

$H^0(\tilde{X}, K_{\tilde{X}})=H^0(\sigma^* K_X)=H^0(K_X)$,

the second equality following from the fact that $\sigma$ has degree $1$.

Since the computation is essentially the same, let's do it in the $2$-dimensional case. So let us assume first $X=\mathbb{A}^2$ with coordinates $x, y$. We have

$\tilde{X}=\{\lambda x+ \mu y=0\} \subset \mathbb{A}^2 \times \mathbb{P}^1$.

In the affine chart $\mu=1$ the equation is therefore $y=\lambda x$, and the exceptional divisor $E$ is given by $x=0$.

Now let $\omega$ be a meromorphic $2$-form on $X$, that we can write as

$\omega=\frac{f(x,y)}{g(x,y)}dx \wedge dy$.

Therefore

$\sigma^* \omega=\frac{f(x, \lambda x)}{g(x, \lambda x)} dx \wedge d(\lambda x)= x \frac{f(x, \lambda x)}{g(x, \lambda x)} dx \wedge d \lambda$.

Therefore in a neighborhood of $E$ we have

$\textrm{div} \ (\sigma^* \omega)= \sigma^* (\textrm{div} \ \omega) + E$.

Since clearly $\textrm{div} \ (\sigma^* \omega)= \sigma^* (\textrm{div} \ \omega)$ outside $E$, it follows in particular that the divisor of every holomorphic form on $\tilde{X}$ contains the component $x=0$, i.e. the exceptional divisor.

Now take $X$ projective. The previous local computation shows that the 
exceptional divisor is contained in the fixed part of the canonical system of $\tilde{X}$, namely

$|K_{\tilde{X}}|= E + |K_X|$

(if you prefer, this also follows from $K_{\tilde{X}} E=-1$).

This implies

$H^0(\tilde{X}, K_{\tilde{X}})=H^0(\sigma^* K_X)=H^0(K_X)$,

the second equality following from the fact that $\sigma$ has degree $1$.

Since the computation is essentially the same, let's do it in the $2$-dimensional case. The problem being local, we may assume $X=\mathbb{A}^2$ with coordinates $x, y$. So

$\tilde{X}=\{\lambda x+ \mu y=0\} \subset \mathbb{A}^2 \times \mathbb{P}^1$.

In the affine chart $\mu=1$ the equation is therefore $y=\lambda x$, and the exceptional divisor $E$ is given by $x=0$.

Now let $\omega$ be a meromorphic $2$-form on $X$, that we can write as

$\omega=\frac{f(x,y)}{g(x,y)}dx \wedge dy$.

Therefore

$\sigma^* \omega=\frac{f(x, \lambda x)}{g(x, \lambda x)} dx \wedge d(\lambda x)= x \frac{f(x, \lambda x)}{g(x, \lambda x)} dx \wedge d \lambda$.

Therefore in a neighborhood of $E$ we have

$\textrm{div} \ (\sigma^* \omega)= \sigma^* (\textrm{div} \ \omega) + E$.

Since clearly $\textrm{div} \ (\sigma^* \omega)= \sigma^* (\textrm{div} \ \omega)$ outside $E$, it follows

$K_{\hat{X}}=\sigma^* K_{X} + E$.

Since the computation is essentially the same, let's do it in the $2$-dimensional case. So let us assume $X=\mathbb{A}^2$ with coordinates $x, y$. We have

$\tilde{X}=\{\lambda x+ \mu y=0\} \subset \mathbb{A}^2 \times \mathbb{P}^1$.

In the affine chart $\mu=1$ the equation is therefore $y=\lambda x$, and the exceptional divisor $E$ is given by $x=0$.

Now let $\omega$ be a meromorphic $2$-form on $X$, that we can write as

$\omega=\frac{f(x,y)}{g(x,y)}dx \wedge dy$.

Therefore

$\sigma^* \omega=\frac{f(x, \lambda x)}{g(x, \lambda x)} dx \wedge d(\lambda x)= x \frac{f(x, \lambda x)}{g(x, \lambda x)} dx \wedge d \lambda$.

Therefore in a neighborhood of $E$ we have

$\textrm{div} \ (\sigma^* \omega)= \sigma^* (\textrm{div} \ \omega) + E$.

Since clearly $\textrm{div} \ (\sigma^* \omega)= \sigma^* (\textrm{div} \ \omega)$ outside $E$, it follows in particular that the divisor of every holomorphic form on $\tilde{X}$ contains the component $x=0$, i.e. the exceptional divisor. In other words, the exceptional divisor is contained in the fixed part of the canonical system of $\tilde{X}$, namely

$|K_{\tilde{X}}|= E + |K_X|$.

This implies

$H^0(\tilde{X}, K_{\tilde{X}})=H^0(\sigma^* K_X)=H^0(K_X)$,

the second equality following from the fact that $\sigma$ has degree $1$.