The Channel Impulse Response (CIR) for a beamforming vector $\mathbf{w} \in \mathbb{C}^{N_T \times 1}$ (which corresponds to your spatial streams) is the following
$$\mathbf{c}(n) = \sum\limits_{k=1}^q \alpha_k \mathbf{a}_{R,k}\mathbf{a}_{T,k}^\top \mathbf{w} \delta(n - \tau_k) \tag{1}$$
where $q$ is the number of paths (is $2$ due to the two-ray propagation model), $\alpha_k$ is the complex gain of the $k^{th}$ path, $ \mathbf{a}_{R,k}$ and $\mathbf{a}_{T,k}$ are the receive and transmit steering vectors of the $k^{th}$ path (depending on the antenna placement and the Angles-of-Arrival and Angles-of-Departures of the path) and finally $\tau_k$ is the propagation delay of the $k^{th}$ path.
I can see that in your case $\mathbf{a}_{T,k} \in \mathbb{C}^{2 \times 1},\mathbf{a}_{R,k} \in \mathbb{C}^{2 \times 1},\mathbf{w} \in \mathbb{C}^{2 \times 1}$. Note that equation $1$ could be written as
$$\mathbf{c}(n) = \Big(\sum\limits_{k=1}^q \alpha_k \mathbf{a}_{R,k}\mathbf{a}_{T,k}^\top \delta(n - \tau_k) \Big) \mathbf{w}$$
So in order to "equalize" to get $\mathbf{w}$, you need to know the matrix $\mathbf{H} = \sum\limits_{k=1}^q \alpha_k \mathbf{a}_{R,k}\mathbf{a}_{T,k}^\top \delta(n - \tau_k) $.