We first show that
$$\langle \{\vec{v}_1, \vec{v}_2 \} \rangle \subset \langle \{2\vec{v}_1 - \vec{v}_2, \vec{v}_1 + 2\vec{v}_2 \} \rangle.$$
To that end, suppose that $a_1$ and $a_2$ are scalars so that
$$a_1\vec{v}_1 + a_2\vec{v}_2 \in \langle \{\vec{v}_1, \vec{v}_2 \} .$$
We must find scalars $b_1$ and $b_2$ such that
$$a_1\vec{v}_1 + a_2\vec{v}_2 = b_1 (2\vec{v}_1 - \vec{v}_2) + b_2 (\vec{v}_1 + 2\vec{v}_2).$$
Rewriting the right hand side, we find
$$a_1\vec{v}_1 + a_2\vec{v}_2 = (2 b_1 + b_2)\vec{v}_1 + (-b_1 + 2b_2)\vec{v}_2.$$
Thus, finding $b_1$ and $b_2$ is equivalent to solving the linear system
$$
\begin{align}
2b_1 + b_2 &= a_1 \\
-b_1 + 2b_2 &= a_2.
\end{align}
$$
Of course, this has a solution as long as the coefficient matrix
$$
\left(\begin{matrix}
2 & 1 \\
-1 & 2
\end{matrix}\right)
$$
is nonsingular. Of course, the matrix *is* nonsingular, since the columns are linearly independent. (See theorem NME3 on page 161.) Note that the reverse inclusion can be obtained immediately by applying the inverse matrix!
Again, we first show that
$$\langle \{\vec{v}_1, \vec{v}_2 \} \rangle \subset \langle \{2\vec{v}_1 - \vec{v}_2, \vec{v}_1 + 2\vec{v}_2 \} \rangle.$$
So, suppose that $a_1$ and $a_2$ are scalars so that
$$a_1\vec{v}_1 + a_2\vec{v}_2 \in \langle \{\vec{v}_1, \vec{v}_2 \} .$$
We then set
$$
\begin{align}
b_1 &= (2a_1-a_2)/5 \\
b_2 &= (a_1+2a_2)/5.
\end{align}
$$
Then
$$
\begin{align}
2b_1+b_2 &= \frac{2}{5}(2a_1-a_2) + \frac{1}{5}(a_1+2a_2) = a_1\\
-b_1+2b_2 &= \frac{1}{5}(-2a_1+a_2) + \frac{2}{5}(a_1+2a_2) = a_2.
\end{align}
$$
Thus,
$$
b_1 (2\vec{v}_1 - \vec{v}_2) + b_2 (\vec{v}_1 + 2\vec{v}_2) = (2 b_1 + b_2)\vec{v}_1 + (-b_1 + 2b_2)\vec{v}_2 = a_1\vec{v}_1 + a_2\vec{v}_2.
$$
Note that the reverse inclusion is a bit simpler because, as we can see reading the last equation from right to left, given $b_1$ and $b_2$ we simply take
$$
\begin{align}
a_1 &= 2b_1+b_2 \\
a_2 &= -b_1+2b_2.
\end{align}
$$
Comments
Proof 1
We first show that
$$\langle \{\vec{v}_1, \vec{v}_2 \} \rangle \subset \langle \{2\vec{v}_1 - \vec{v}_2, \vec{v}_1 + 2\vec{v}_2 \} \rangle.$$
To that end, suppose that $a_1$ and $a_2$ are scalars so that
$$a_1\vec{v}_1 + a_2\vec{v}_2 \in \langle \{\vec{v}_1, \vec{v}_2 \} .$$
We must find scalars $b_1$ and $b_2$ such that
$$a_1\vec{v}_1 + a_2\vec{v}_2 = b_1 (2\vec{v}_1 - \vec{v}_2) + b_2 (\vec{v}_1 + 2\vec{v}_2).$$
Rewriting the right hand side, we find
$$a_1\vec{v}_1 + a_2\vec{v}_2 = (2 b_1 + b_2)\vec{v}_1 + (-b_1 + 2b_2)\vec{v}_2.$$
Thus, finding $b_1$ and $b_2$ is equivalent to solving the linear system
$$
\begin{align}
2b_1 + b_2 &= a_1 \\
-b_1 + 2b_2 &= a_2.
\end{align}
$$
Of course, this has a solution as long as the coefficient matrix
$$
\left(\begin{matrix}
2 & 1 \\
-1 & 2
\end{matrix}\right)
$$
is nonsingular. Of course, the matrix *is* nonsingular, since the columns are linearly independent. (See theorem NME3 on page 161.) Note that the reverse inclusion can be obtained immediately by applying the inverse matrix!
Proof 2
Again, we first show that
$$\langle \{\vec{v}_1, \vec{v}_2 \} \rangle \subset \langle \{2\vec{v}_1 - \vec{v}_2, \vec{v}_1 + 2\vec{v}_2 \} \rangle.$$
So, suppose that $a_1$ and $a_2$ are scalars so that
$$a_1\vec{v}_1 + a_2\vec{v}_2 \in \langle \{\vec{v}_1, \vec{v}_2 \} .$$
We then set
$$
\begin{align}
b_1 &= (2a_1-a_2)/5 \\
b_2 &= (a_1+2a_2)/5.
\end{align}
$$
Then
$$
\begin{align}
2b_1+b_2 &= \frac{2}{5}(2a_1-a_2) + \frac{1}{5}(a_1+2a_2) = a_1\\
-b_1+2b_2 &= \frac{1}{5}(-2a_1+a_2) + \frac{2}{5}(a_1+2a_2) = a_2.
\end{align}
$$
Thus,
$$
b_1 (2\vec{v}_1 - \vec{v}_2) + b_2 (\vec{v}_1 + 2\vec{v}_2) = (2 b_1 + b_2)\vec{v}_1 + (-b_1 + 2b_2)\vec{v}_2 = a_1\vec{v}_1 + a_2\vec{v}_2.
$$
Note that the reverse inclusion is a bit simpler because, as we can see reading the last equation from right to left, given $b_1$ and $b_2$ we simply take
$$
\begin{align}
a_1 &= 2b_1+b_2 \\
a_2 &= -b_1+2b_2.
\end{align}
$$