So, since $V\in\mathbb R^{m\times n}$ has full column rank $n$, implying also $m\geq n$, we might obtain the pseudoinverse using
$$
V^+=(V^TV)^{-1}V^T
$$
since $V^TV$ is invertible and $V^*=V^T$ in $\mathbb R$.
Now, let's denote by $i$ the index of the row which is scaled and by $d_i$ - the scaling factor.
Thus, a perturbed matrix $V^\prime$ can be expressed as
$$
V^\prime = D_i V= \text{diag}(\underbrace{1,\ldots,1}_{i-1},d_i,\underbrace{1,\ldots,1}_{m-i})V
$$
Now, how can we express it as a low-rank update?
Consider
$$
u=e_i=\left[
\begin{array}{c}
0\\
\vdots\\
1 \\
\vdots\\
0
\end{array}\right], \quad\quad
v=(V_{i,*})^T
$$
where $u$ is the $i$th elementary unit vector in $m$ dimensional space and $v$ is the transposed $i$th row of the original matrix $V$. Then:
$$
V^\prime = V+(d_i-1)uv^T
$$
Now, let's express the update to the "correlation" matrix $V^T V$:
$$
\begin{aligned}
{V^\prime}^T{V^\prime}&=(V+(d_i-1)uv^T)^T(V+(d_i-1)uv^T)\\
&=V^TV+(d_i-1)v\underbrace{u^TV}_{v^T}+(d_i-1)\underbrace{V^Tu}_{v}v^T+(d_i-1)(d_i-1)v\underbrace{u^Tu}_{1}v^T\\
&=V^TV+\left((d_i-1)+(d_i-1)+(d_i-1)^2\right)vv^T\\
&=V^TV+(d_i^2-1)vv^T
\end{aligned}
$$
With that, we expressed a rank-1 update to the correlation matrix $V^T V$ for which we already know the inverse. For compactness, let's denote
$$\gamma=d_i^2-1$$.
Now, let's use Sherman-Morrison to find an update to the inverse of the correlation matrix:
$$
\begin{aligned}
({V^\prime}^T{V^\prime})^{-1}&=(V^TV+\gamma v v^T)^{-1}\\
&=(V^TV)^{-1}-\frac{\gamma (V^TV)^{-1}vv^T(V^TV)^{-1}}{1+\gamma v^T (V^TV)^{-1}v}
\end{aligned}
$$
Let's denote the following matrix-vector product by
$$
w = (V^TV)^{-1}v
$$
Then, since $V^T V$ is symmetric
$$
\begin{aligned}
({V^\prime}^T{V^\prime})^{-1}&=(V^TV)^{-1}-\frac{\gamma ww^T}{1+\gamma v^T w}\\
&=(V^TV)^{-1}-\tau ww^T, \quad \tau=\frac{d_i^2-1}{1+(d_i^2-1)v^Tw}
\end{aligned}
$$
So, in order to update the inverse of the correlation matrix (provided you have the computation for the original matrix done):
- Compute matrix-vector product $w = (V^TV)^{-1}v$ in $n^2$ operations.
- Calculate dot-product of $w$ and original row $v$: $v^T w$ in $n$ operations.
- Calculate $\tau$ in constant time.
- Calculate the scaled by $\tau$ outer product $\tau ww^T$ in $n^2$ operations
- Calculate the update to the inverse of the correlation matrix by subtracting the outer product in $n^2$ operations.
So, you achieved $({V^\prime}^T{V^\prime})^{-1}$ in $3n^2+n$ operations. Probably, can be done in $2n^2+n$ if fused flops are used combining steps 4 and 5.
Now, you can calculate
$$
{V^\prime}^+ = ({V^\prime}^T{V^\prime})^{-1}V^TD_i
$$
or leave it in the unassembled form depending on your future usage.
I am not sure if that will lead to significant speed-up for the requested $\mathbb R^{4\times 3}$ case but is worth trying.
Note, this particular technique relies on $V$ having full-column rank, otherwise, the pseudoinverse cannot be found this way and SVD decomposition has to be used -> where updates are also possible, but much trickier.