Theoretical reason for multiple linear regression predictions being the same when adding and subtracting predictors Say I have two variables $x_1$ and $x_2$, now I build a linear regression model as below
$$\hat{Y} = n_1 x_1 + n_2 x_2.$$
Then I build another model as below
$$\hat{Z} = m_1 (x_1 + x_2) + m_2 (x_1 - x_2).$$
Intuitively, $\hat{Y}$ should be equal to $\hat{Z}$.  Below is my R code to demonstrate the equivalence
set.seed(1) 
num = 10
X1 = runif(num)
X2 = runif(num)
Y = runif(num)

mydata <- data.frame(X1, X2, Y)
fit1 = lm(Y ~ X1 + X2, data = mydata)
summary(fit1)

mydata <- data.frame(X1 + X2, X1 - X2, Y)
names(mydata)[1] <- 'new_X1'
names(mydata)[2] <- 'new_X2'

fit2 = lm(Y ~ new_X1 + new_X2, data = mydata)
summary(fit2)

My questions is, how can I conceptually prove the equivalence?
 A: This occurs because of the properties of the hat matrix
The theoretical reason that this occurs is due to the properties of a projection matrix in linear regression called the "hat matrix".  In linear regression, the predicted response vector under OLS estimation is:
$$\hat{\mathbf{y}} = \mathbf{h} \mathbf{y}
\quad \quad \quad \quad \quad 
\mathbf{h} \equiv \mathbf{h}(\mathbf{x}) \equiv \mathbf{x} (\mathbf{x}^\text{T} \mathbf{x})^{-1} \mathbf{x}^\text{T}.$$
The matrix $\mathbf{h}$ (called the "hat matrix") is a projection matrix that projects onto the column-space of the design matrix $\mathbf{x} =[\mathbf{x}_1,...,\mathbf{x}_k]$.  This column-space is the set of values:
$$\mathscr{C}(\mathbf{x}) = \bigg\{ \sum_{i=1}^k \omega_i \mathbf{x}_i \bigg| \omega_1,...,\omega_k \in \mathbb{R} \bigg\}.$$
It can easily be shown that if you make an adjustment to the design matrix by a linear transformation (with no zero weights) then the column-space is unchanged.  In your particular example, the two formulations of the explanatory variables in your two models lead to design matrices that are linear transformations of one another, and most importantly, they have the same column-space.  Because the hat matrix is a projection matrix that projects onto the column-space of the design matrix, you have the implication:
$$\mathscr{C}(\mathbf{x}) = \mathscr{C}(\mathbf{x}_*)
\quad \quad \implies \quad \quad
\mathbf{h}(\mathbf{x}) = \mathbf{h}(\mathbf{x}_*).$$
In particular, if we create a new design matrix $\mathbf{x}_* = \mathbf{x} \mathbf{A}$ using some invertable linear transformation $\mathbf{A}$ (which is a $k \times k$ matrix in the present context) then we will have $\mathscr{C}(\mathbf{x}) = \mathscr{C}(\mathbf{x}_*)$ and $\mathbf{h}(\mathbf{x}) = \mathbf{h}(\mathbf{x}_*)$ (see direct proof of the latter in the section below).
Now, in your question you use linear regression models for two different sets of explanatory variables, where the resulting design matrices are related by an invertible linear transformation.  This means that while your design matrices are different, they have the same column-space and they have the same hat matrix.  Consequently, the predicted response vector in each case is the same.

Invariance of hat matrix to invertible linear transformation of design: Suppose we create a new design matrix $\mathbf{x}_* = \mathbf{x} \mathbf{A}$ using some invertable linear transformation $\mathbf{A}$.  The hat matrix for the new design matrix is:
$$\begin{align}
\mathbf{h}(\mathbf{x}_*)
&= \mathbf{h}(\mathbf{x} \mathbf{A}) \\[6pt]
&= (\mathbf{x} \mathbf{A}) ((\mathbf{x} \mathbf{A})^\text{T} (\mathbf{x} \mathbf{A}))^{-1} (\mathbf{x} \mathbf{A})^\text{T} \\[6pt]
&= \mathbf{x} \mathbf{A} (\mathbf{A}^\text{T} (\mathbf{x} ^\text{T} \mathbf{x}) \mathbf{A})^{-1} \mathbf{A}^\text{T} \mathbf{x}^\text{T} \\[6pt]
&= \mathbf{x} \mathbf{A} \mathbf{A}^{-1} (\mathbf{x} ^\text{T} \mathbf{x})^{-1} (\mathbf{A}^\text{T})^{-1} \mathbf{A}^\text{T} \mathbf{x}^\text{T} \\[6pt]
&= \mathbf{x} (\mathbf{x} ^\text{T} \mathbf{x})^{-1} \mathbf{x}^\text{T} \\[6pt]
&= \mathbf{h}(\mathbf{x}). \\[6pt]
\end{align}$$
