Proof regarding partial derivatives and differentiability

I'm having a trouble in understanding the proof of Theorem 8 from Pugh's Real Mathematical Analysis 2nd Edition (p. 284).On the 5th line from the bottom where use has been made of the chain rule and mean value theorem, I don't understand why the last term doesn't end with a $v_je_j$ instead of just $v_j$:$$f_i(p_j) − f_i(p_{j−1}) = g(1) − g(0) = g'(t_{ij}) = \frac{\partial f_i(p_{ij})}{\partial x_j} v_j.\label{1}\tag{1}$$The reason I think the $e_j$ is missing is based on the following.
The proof states:

$$\sigma_j(t) = p_{j−1} + tv_je_j, \quad0 ≤ t ≤ 1$$By the one-dimensional chain rule and mean value theorem applied to the differentiable real-valued function $g(t) = f_i \circ \sigma_j(t)$ of one variable, there exists $t_{ij} ∈ (0, 1)$ such that$$f_i(p_j) − f_i(p_{j−1}) = g(1) − g(0) = g'(t_{ij}) = \frac{\partial f_i(p_{ij})}{\partial x_j} v_j$$where $p_{ij} = \sigma_j(t_{ij})$.

So, in order to get $g'(t_{ij})$ we apply the chain rule which results in the term $\dfrac{\partial f_i(p_{ij})}{\partial x_j}$ multiplied by the derivative of $\sigma_j(t_{ij})$.
I see that the derivative of the function $\sigma_j(t) = p_{j−1} + tv_je_j$ with respect to $t$ is $v_je_j$ so we end up with the following expression:$$g^\prime(t_{ij})=\frac{\partial f_i(p_{ij})}{\partial x_j}v_je_j$$whereas the book does not show the $e_j$. If someone is able to correct my reasoning, that would be great.

My understanding is as follows:

• $f_i$ takes in a vector and spit out a scalar,
• $v$ is a vector,
• $v_j$ is a scalar $e_j$ is a basis vector
• $σ_j(t)$ takes in a scalar and spits out a vector
• $\frac{\partial f_i(p_{ij})}{\partial x_j}$ is a scalar
• $p_{ij} = σ_j(t_{ij})$ is a vector
• $p$ is a vector

Hence, the LHS of \eqref{1}, i.e. $f_i(p_j) − f_i(p_{j−1})$ is a scalar, so the RHS needs to be the same way, but if the $e_j$ is present I think that would make the RHS term, i.e.$$\frac{\partial f_i(p_{ij})}{\partial x_j} v_je_j,$$a vector which won't work.

Looking for clarification on where I'm messing up.

And while I'm here, the next bits of the proof introduce $R_i(v)$, seeking clarification of whether this is a scalar, seems it should be.