Remember when you solved systems of linear equations in school? This is essentially the same. You just swap out $K=\mathbb R$ for $K=\mathbb F_{601}$ which allows you to essentially do the same things as with $\mathbb R$, even if it is a little bit less inttuitive.
First too $\mathbb F_{601}$, as $p=601$ is prime, this is what is called a field. Essentially instead of doing normal addition, you do normal addition first and then reduce the result $\bmod p$ (ie take the remainder after integer division by $p$), same for subtraction and multiplication. Exponentiation works the same (as it is just repeated multiplication) and you can use the identity $1/a:=a^{p-2}\bmod p$. Now you can view your problem to solve the following equation over this field:
$$\begin{pmatrix}
m_1&1\\
m_2&1
\end{pmatrix}
\cdot
\begin{pmatrix}
k_1\\k_2
\end{pmatrix}
=
\begin{pmatrix}
c_1\\c_2
\end{pmatrix}$$
Obviously you will bring the matrix into triangle form (by subtracting the first line from the second) resulting in
$$\begin{pmatrix}
m_1&1\\
m_2-m_1&0
\end{pmatrix}
\cdot
\begin{pmatrix}
k_1\\k_2
\end{pmatrix}
=
\begin{pmatrix}
c_1\\c_2-c_1
\end{pmatrix}$$
Implying that $k_1=\frac{c_2-c_1}{m_2-m_1}$, which is indeed expected if we consider the cipher to be a linear function like those you learned in school. Now it immediately follows that $k_1\cdot m_1+k_2=c_1\iff k_2=c_1-k_1\cdot m_1$ and you have thus successfully recovered the full key $\begin{pmatrix}
k_1\\k_2
\end{pmatrix}$.
I'll leave it as an exercise to plug-in the numbers and recover the concrete values using a calculator or some online tool.
