You can also tensor product the operations you are applying to your bipartite system. Suppose you have $X$ on the first qubit and nothing on the second qubit, i.e., the identity gate $I$. Thus, to get the matrix you multiply your four-vector by, you tensor product these operators as $X \otimes I$ (or $I \otimes X$ if you are following little endian convention). This tensor products works as usual:
$$
X \otimes I = \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix} \otimes \begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix} = \begin{bmatrix}0\begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix}&1\begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix} \\ 1\begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix}&0\begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix}\end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\end{bmatrix}
$$
Then, you can multiply this matrix times your column vector with four elements. Perhaps this is easier to do following ket notation, particularly if you have more than two qubits. For example, assuming you have the $\frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)$ state, you can apply $X \otimes I$ as follows:
$$
\begin{align}
X \otimes I \left[\frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)\right] &= \frac{1}{\sqrt{2}}\left[ X\otimes I(|01\rangle + |10\rangle) \right] \\ &=\frac{1}{\sqrt{2}}(X|0\rangle I|1\rangle + X|1\rangle I|0\rangle) = \frac{1}{\sqrt{2}}(|11\rangle + |00\rangle)
\end{align}
$$
