As I mentioned in the long comment chain under OP, assuming $y_{u,i}$ is scalar, $v_i$ is a row vector and $x_u$ is a column vector (with matching sizes), we can show that:
$$L=\sum_{u,i}(y_{u,i}-v_ix_u)^2+\lambda\left(\sum_i\|v_i\|_2^2+\sum_u\|x_u\|_2^2\right)$$
$$\frac{\partial L}{\partial x_a}
=\frac{\partial }{\partial x_a}\left\{\sum_{i}(y_{a,i}-v_ix_a)^2+\lambda\|x_a\|_2^2\right\}\\
=\frac{\partial }{\partial x_a}\left\{\sum_{i}(y_{a,i}-v_ix_a)^2\right\}+\lambda\frac{\partial }{\partial x_a}\left\{\|x_a\|_2^2\right\}\\
=-2\sum_{i}(y_{a,i}-v_ix_a)v_i^T+2\lambda x_a\\
$$
By similar analogy
$$\frac{\partial L}{\partial v_j}
=-2\sum_{u}(y_{u,j}-v_jx_u)x_u^T+2\lambda v_j\\
$$
Notice that both gradients were assumed (by me) to have the same dimensions of the parameters.
Bar any error, you were missing only the derivative of the squared $\ell_2$ norm of a vector. This can be simply be shown to be, for vector $w$ (irrespective of the nature of $w$, be it a column or row matrix).
$$\frac{\partial \|w\|_2^2}{\partial w}=2w$$
Addendum: if you take vector derivatives to have a specific configuration, e.g. always rows or always columns, then some adjustments are due. Since you did not assume vectors to be column matrices (as we can see by the vector product), I think my solution is on point.
